Nuclei: Class 12 Physics Chapter 13 Hinglish mein

Physics Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

90 min
16 practice questions
29 MCQs

Is chapter mein aap kya seekhenge

Nucleus ki banawat ko Z, N aur A se describe karo, atomic mass unit use karo, aur nuclides ko isotopes, isobars ya isotones mein classify karo
Samjho ki scattering experiments se nucleus ka size kaise tay hota hai, R = R0 A^(1/3) use karo aur dikhao ki nuclear density sabhi nuclei ke liye same hai
Atomic masses se mass defect, binding energy aur binding energy per nucleon nikaalne ke liye E = mc² apply karo
Binding-energy curve ko samjho aur usse saturation, fission aur fusion explain karo
Nuclear force ki main properties batao aur uska potential-energy curve padho
Alpha, beta aur gamma decay pehchano, aur har ek Z aur A mein kya badlav laata hai, yeh batao
Nuclear reactions ke Q-values calculate karo aur nuclear aur chemical energy ko compare karo
Samjho ki fission aur thermonuclear fusion energy kaise release karte hain, aur fusion ko bahut high temperature kyun chahiye

NCERT Physics Class 12, Chapter 13, pp. 306–322 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

13.1 Introduction

Tumhare chehre ko garam karti sunlight Sun ke core se us energy ke roop mein nikli thi jo hydrogen nuclei ke judkar helium banane par free hui. Tumhare ghar ki kuch electricity shayad Kudankulam ya Tarapur se aati ho, jahan uranium nuclei split hote hain. Dono tarah ki energy nucleus (नाभिक) se aati hai, atom ka woh chhota sa dil jise pichhle chapter mein Rutherford ke alpha-particle experiments ne dhoondh nikala tha.

Yeh kitna chhota hai? Ek typical atom lagbhag $10^{- 10} m$ across hota hai, jabki uska nucleus lagbhag $10^{- 14} m$ : radius mein das hazaar guna chhota. Volume radius ke cube ke saath badhta hai, isliye nucleus atom ke volume ka sirf lagbhag $(10^{- 4})^{3} = 10^{- 12}$ hissa bharta hai. Socho ek atom ko cricket stadium jitna bada kar diya gaya: nucleus centre spot par rakha ek angoor hoga, aur electrons stands mein ghoom rahe honge. Phir bhi woh angoor lagbhag saara mass rakhta hai. Hydrogen atom mein electron mass ka sirf 0.05 % hai; bhaari atoms mein electrons ka hissa aur bhi kam hai, isliye har atom mein nucleus total mass ka 99.9 % se zyada rakhta hai.

3/3Nitrogen-14 nucleus mein 7 protons (+) aur 7 neutrons (n) hote hain.

Fig. 13.1Das hazaar guna zoom karne par atom ke centre ka dot protons aur neutrons se bana nucleus ban jaata hai.NCERT p. 306

Step 1 mein atom ek bada dashed sky-blue cloud hai, jismein teen chhote electrons hain aur centre mein ek akela dark dot hai, jis par nucleus likha hai; atom lagbhag ten to the minus ten metres across hai. Step 2 zoom in karta hai: cloud phoolkar fade ho jaata hai, aur dot chaudah balls ka ek gol cluster ban jaata hai, jis par ab nucleus likha hai, lagbhag ten to the minus fourteen metres across. Step 3 har ball ko mark karta hai: saat protons par plus sign hai aur saat neutrons par letter n, saath mein ek key aur label: nitrogen-14, saat protons plus saat neutrons.

Yeh chapter us angoor ke baare mein chaar sawaal poochhta hai. Yeh kis cheez ka bana hai, aur hum iski banawat kaise likhte hain? Yeh kitna bada hai, aur kitna dense? Yeh tika hi kyun rehta hai, jab iske protons ek doosre ko repel karte hain? Aur nuclei ko rearrange karke power stations aur stars jitni energy kaise release ho sakti hai? Raaste mein hum radioactivity, fission aur fusion se bhi milenge.

Quick check

2 mein se 1

Nucleus ka radius uske atom ke radius ka lagbhag

10^{- 4}

guna hai. Nucleus atom ke volume ka lagbhag kitna hissa gherta hai?

10^{- 4}

10^{- 8}

10^{- 12}

10^{- 16}

Answer dekho

Answer (C) $10^{- 12}$

Volume radius ke cube ke saath scale hota hai:

(10^{- 4})^{3} = 10^{- 12}

10^{- 4}

chunne ka matlab hai yeh bhool jaana ki sphere ka volume

R^{3}

par depend karta hai,

R

par nahi.

13.2 Atomic masses aur nucleus ki composition

Atoms ke liye kilogram ek bhadda unit hai: ek carbon atom ka mass lagbhag $2 \times 10^{- 26} k g$ hai. Isliye physicists aur chemists atomic mass unit (symbol u) use karte hain, jo carbon-12 ke ek atom ke mass ka barahvaan hissa define kiya gaya hai. Iski value kilograms mein hum mole se nikaal sakte hain. Mass unit ki definition se, carbon-12 atoms ke ek mole ka mass 12 g hai, aur ek mole mein Avogadro's number $N_{A} = 6.022 \times 10^{23}$ atoms hote hain. To ek carbon-12 atom ka mass $0.012 k g / N_{A} = 1.9926 \times 10^{- 26} k g$ hai, aur uska barahvaan hissa atomic mass unit hai.

Atomic mass unit

1 u = \frac{m (^{12} C)}{12} = 1.660539 \times 10^{- 27} k g

(13.1)

Definition

Atomic mass unit (u)

Ek carbon-12 atom ke mass ka barahvaan hissa: $1 u = 1.6605 \times 10^{- 27} k g$ .

ExampleEk hydrogen atom ka mass 1.007825 u hai.

u mein naapne par atoms ke masses lagbhag whole numbers aate hain, jaise har atom lagbhag hydrogen atom jitne mass ke blocks se bana ho. Kuch elements is pattern ko buri tarah todte hain: copper 63.55 u aata hai aur magnesium 24.31 u. Iska explanation mass spectrometer se mila, jo ions ko unke charge-to-mass ratio se alag karta hai aur is tarah ek-ek atom ko tolta hai. Isne dikhaya ki zyaadatar elements aise atoms ka mixture hain jinki chemistry same hai lekin masses alag. Inhe isotopes (समस्थानिक) kehte hain, Greek words se jinka matlab hai same place, kyunki yeh periodic table mein ek hi jagah share karte hain. Periodic table mein chhapa atomic mass isotope masses ka average hai, jismein har isotope ko uske common hone ke hisaab se weight diya jaata hai (uski relative abundance).

Example

Boron 10 ya 11 kyun nahi hai

Natural boron mein 19.9 % boron-10 (atomic mass 10.0129 u) aur 80.1 % boron-11 (atomic mass 11.0093 u) hai. Boron ka average atomic mass nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ek ordinary hydrogen atom se uska akela electron hata do, to jo bachta hai woh ek bare proton hai. Isliye proton ka mass do measured numbers se nikalta hai, hydrogen atom ka mass (1.007825 u) aur electron ka mass, $m_{e} = 0.000549 u$ :

Proton ka mass

\begin{aligned} m_{p} & = 1.007825 u \\ - 0.000549 u \\ = 1.007276 u \\ = 1.67262 \times 10^{- 27} k g \end{aligned}

(13.2)

Nature mein hydrogen ka 99.985 % ordinary hydrogen hai. Baaki lagbhag saara deuterium (atomic mass 2.014102 u) hai, jo paani ke har sample mein milta hai. Teesra isotope, tritium (3.016049 u), unstable hai: kisi bhi sample ka aadha hissa lagbhag 12 saal mein decay ho jaata hai, isliye labs mein use hone wala tritium banaya jaata hai.

Ek nucleus mein kitne protons hote hain? Charges gino. Atomic number $Z$ wale neutral atom ke charon taraf $Z$ electrons hote hain, jin par kul $- Z e$ charge hai, isliye uske nucleus par $+ Z e$ hona chahiye. Har proton par $+ e$ hai, isliye nucleus mein exactly $Z$ protons hote hain. Kya kuch electrons nucleus ke andar chhupe ho sakte hain, jo extra protons ka charge cancel kar dein? Physicists ko kabhi aisa shak tha, lekin quantum mechanics ise mana karti hai: kuch femtometres ke region mein dabaye gaye electron ki energy itni zyada hogi ki koi bhi nucleus use pakad kar nahi rakh sakta.

Neutron ki discovery

Hydrogen isotopes ne ek paheli khadi kar di. Teeno nuclei ka charge $+ e$ hai, isliye har ek mein sirf ek proton hai, phir bhi unke masses lagbhag 1 : 2 : 3 ke ratio mein hain. Isliye deuterium aur tritium nuclei mein extra neutral matter hona chahiye, lagbhag ek aur do proton masses jitna. 1932 mein James Chadwick ne pehchana ki woh neutral matter kya hai (Fig. 13.2).

4/4Energy aur momentum tabhi balance hote hain jab yeh rays lagbhag proton jitne mass wale neutral particles hon.

Fig. 13.2Beryllium par girte alpha particles ek neutral radiation nikaalte hain jo paraffin wax se tez protons bahar nikaal deta hai; un protons se ulta calculate karke neutron ka pata chala.

Left se right chaar objects ki ek row: ek box jis par alpha source likha hai, ek patli purple slab jis par beryllium likha hai, ek moti pale-yellow slab jis par paraffin likha hai, aur needle wala ek box jis par counter likha hai. Step 1: ek arrow alpha particles ko source se beryllium tak le jaata hai. Step 2: ek dashed arrow, jis par likha hai ki neutral rays E ya B fields se mudti nahi, beryllium se paraffin tak jaata hai. Step 3: ek solid arrow fast protons ko, jo plus sign wali ball se dikhaye gaye hain, paraffin se counter tak le jaata hai, aur counter ki needle ghoom jaati hai. Step 4: ek highlighted card par n mark ki hui ball hai aur likha hai: neutral, mass lagbhag proton jitna.

Pehle ke experimenters ne dekha tha ki alpha particles se bombard kiya gaya beryllium ek bahut penetrating, uncharged radiation deta hai, aur yeh radiation paraffin wax se protons ko high speed par bahar maar sakta hai. Unhone ise gamma rays samjha. Chadwick ne dikhaya ki yeh sahi nahi ho sakta. Kisi gamma-ray photon ko proton ko itni kinetic energy dene ke liye, alpha-beryllium collision jitni energy de sakta hai usse kai guna zyada energy chahiye hoti. Lagbhag proton jitna bhaari neutral particle naturally fit baithta hai: ek chalti ball apni energy sabse achhe se same mass wali target ball ko deti hai, jaise billiard balls ki head-on collision mein. Rays se dhakka khaaye protons aur nitrogen nuclei par conservation of energy aur momentum lagakar, Chadwick ne paaya ki naye particle ka mass proton ke mass ke bahut kareeb hai. Unhone ise neutron naam diya, aur is discovery ke liye unhe 1935 ka Physics Nobel Prize mila. Iska mass ab bahut precisely pata hai:

Neutron ka mass

\begin{aligned} m_{n} & = 1.008665 u \\ = 1.674927 \times 10^{- 27} k g \end{aligned}

(13.3)

Neutron proton se thoda bhaari hai. Free neutron zyada der nahi tikta: woh lagbhag 880 s, yaani lagbhag ek chauthai ghante ki mean life ke saath proton, electron aur antineutrino mein badal jaata hai. Stable nucleus ke andar bandha hua wahi neutron hamesha tik sakta hai.

Ab hum kisi bhi nucleus ko uske particles gin kar describe kar sakte hain. Teen whole numbers yeh kaam karte hain: $Z$ protons ko ginta hai (yeh element ka atomic number hai), $N$ neutrons ko ginta hai (neutron number), aur mass number $A$ dono ko saath mein ginta hai:

Mass number

A = Z + N

(13.4)

jahan

$Z$: atomic number, protons ki ginti
$N$: neutron number, neutrons ki ginti
$A$: mass number, nucleons ki total ginti

Definition

Nucleon

Proton ya neutron, woh do tarah ke particles jinse nucleus banta hai. Mass number $A$ nucleons ko ginta hai.

Ek nuclear species, ya nuclide, ko $_{}^{}_{Z}^{A} X$ likhte hain, jahan X chemical symbol hai. Jaise, $_{}^{}_{82}^{208} P b$ lead ka sabse common nucleus hai: 208 nucleons, jinmein se 82 protons hain aur $208 - 82 = 126$ neutrons. Kyunki symbol pehle se $Z$ fix kar deta hai, log aksar subscript hata dete hain aur lead-208 bolte hain ya Pb-208 likhte hain.

Ab isotopes ka simple matlab hai: same $Z$ lekin alag $N$ wale nuclei. Carbon-12, carbon-13 aur carbon-14 sab mein 6 protons hain, aur 6, 7 aur 8 neutrons. Unke atoms ka electron arrangement same hai aur isliye chemistry bhi same, isiliye chemists unhe alag nahi pehchaan paaye. Record tin ke paas hai, das stable isotopes ke saath. Do aur families kaam ki hain. Isobars ka mass number $A$ same hota hai: carbon-14, nitrogen-14 aur oxygen-14 isobars hain jinka $A = 14$ hai. Isotones ka neutron number $N$ same hota hai: carbon-13, nitrogen-14 aur oxygen-15, har ek mein 7 neutrons hain. Ek chart jismein ek axis par $N$ aur doosre par $Z$ ho, teeno families ek nazar mein dikha deta hai (Fig. 13.3).

3/4Isobars ek diagonal par hote hain: same A, yahan 14 (C-14, N-14, O-14).

Fig. 13.3N–Z chart par isotopes ek row share karte hain, isotones ek column share karte hain aur isobars ek diagonal par hote hain.NCERT p. 309

Nau rounded cells ka ek grid, jiske neeche neutron number N = 6, 7 aur 8 hai aur side mein atomic number Z = 6, 7 aur 8. Cells par neeche ki row mein C-12, C-13, C-14, beech ki row mein N-13, N-14, N-15 aur upar ki row mein O-14, O-15, O-16 likha hai; stable nuclides ke borders solid hain aur radioactive walon ke dashed, aur neeche ek key hai. Step 1 khaali chart aur rule A = Z + N dikhata hai. Step 2 neeche ki row ko green mein shade karta hai: isotopes, same Z, alag N. Step 3 O-14 se N-14 hote hue C-14 tak ke diagonal ko orange mein shade karta hai: isobars, same A = 14. Step 4 C-13 se N-14 hote hue O-15 tak beech wale column ko blue mein shade karta hai: isotones, same N = 7.

InteractiveNuclide chart explorer

Free neutron se calcium tak ke light nuclides ka chart, jismein neutron number N right ki taraf aur proton number Z upar ki taraf badhta hai. Stable nuclides solid ink-coloured squares hain aur chune hue radioactive nuclides pale orange squares, saath mein ek dashed N equals Z line. Buttons Z aur N badalte hain, ya tum kisi square par tap kar sakte ho. Cards nuclide ka naam, uska mass number aur R equals 1.2 A to the one-third femtometres se uska radius dikhate hain. Ek status line protons aur neutrons ki ginti, ratio N over Z, aur nucleus stable hai ya kaise decay hota hai, yeh batati hai, aur jab koi highlight button daba ho to uske isotopes, isobars ya isotones list karti hai.

Fig. 13.4Z aur N badal kar light nuclei mein aage-peeche jao, aur jo nucleus chuno uske isotopes, isobars ya isotones highlight karo.

Quick check

4 mein se 1

Iron nucleus

_{}^{}_{26}^{56} F e

mein protons aur neutrons hain.

Answer dekho

Answer (1) 26, (2) 30

Z = 26

protons;

N = A - Z = 56 - 26 = 30

neutrons.

13.3 Nucleus ka size

$10^{- 14} m$ across cheez ko kaise naapoge? Tum use dekh nahi sakte, isliye tum us par cheezein phenkte ho aur dekhte ho ki woh kaise takra kar lautti hain. Pichhle chapter ke Geiger–Marsden experiment mein, gold foil par chalaye gaye alpha particles gold nucleus ke electric repulsion se wapas mud gaye. Rutherford ka formula, jo do point charges ke beech sirf Coulomb's law maanta hai, results se match hua. Yeh success pehle se hi size ke baare mein kuch batati hai: alphas kabhi itne paas nahi aaye ki unhe electric repulsion ke alawa kuch aur mehsoos ho.

Head-on approach mein, alpha particle dheema hota jaata hai kyunki uski kinetic energy $K$ electric potential energy mein badalti hai. Woh distance of closest approach $d$ par ek pal ke liye ruk jaata hai, jahan poori $K$ potential energy ban chuki hoti hai, aur phir wapas mud jaata hai. Alpha particle par charge $2 e$ aur nucleus par $Z e$ ke saath,

Distance of closest approach

\begin{aligned} K & = \frac{1}{4 π ε_{0}} \frac{(2 e) (Z e)}{d} \\ \Rightarrow d & = \frac{1}{4 π ε_{0}} \frac{2 Z e^{2}}{K} \end{aligned}

Kyunki nuclear charge us region se zyada faila nahi ho sakta jismein alpha kabhi ghusa hi nahi, $d$ radius ki upper limit hai (Fig. 13.5). Aise calculations ke liye ek kaam ka number hai $e^{2} / 4 π ε_{0} = 1.44 M e V f m$ , jahan $1 f m = 10^{- 15} m$ (ek femtometre, jise fermi bhi kehte hain).

2/329.5 fm par uski saari kinetic energy electric potential energy ban chuki hai, isliye woh ruk kar wapas mud jaata hai.

Fig. 13.5Kuch MeV wale alphas gold nucleus se kaafi bahar hi wapas mud jaate hain; sirf bahut zyada energetic alphas hi uski surface tak pahunchte hain aur nuclear force mehsoos karte hain.NCERT p. 309

Right mein ek bada orange circle scale ke hisaab se bana gold nucleus hai, jiske centre se left ki taraf ek dashed line jaati hai aur neeche 10 femtometre ka scale bar hai. Step 1: far left par ek chhota purple alpha particle K equal to 7.7 MeV ke saath nucleus ki taraf badhta hai. Step 2: alpha particle us point par ruk gaya hai jahan se woh wapas mudta hai, aur wahan se nucleus ke centre tak ek dimension arrow par d equals 29.5 fm likha hai, jo nucleus se kaafi bahar hai. Step 3: lagbhag 26 MeV K wala ek dashed alpha particle nucleus ki surface ko chhoota dikhaya gaya hai, jis par likha hai ki woh 8.9 fm par chhoota hai, jahan nuclear pull shuru hota hai.

Example

Rutherford ke alphas kitne paas pahunchte hain?

Rutherford ke sabse achhe alpha source ke particles ki kinetic energy 7.7 MeV thi. (a) Head-on collision mein aise alpha particle ka gold nucleus ( $Z = 79$ ) se distance of closest approach nikaalo. (b) Nuclear radii $R = 1.2 A^{1 / 3} f m$ (neeche Eq. 13.5) se lekar batao, gold nucleus ( $A = 197$ ) ko bas chhoone ke liye alpha particle ko kitni energy chahiye hogi?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Tez projectiles ke saath kahani badal jaati hai. Jaise hi projectile nucleus ke edge tak pahunchta hai, short-range nuclear force bhi shamil ho jaata hai aur scattering Rutherford ke formula ko follow karna band kar deta hai; jis energy par yeh deviation shuru hota hai, usse nuclear size milta hai. Isse bhi achhe probes tez electrons hain. Electron nuclear force bilkul mehsoos nahi karta, isliye uska scattering yeh map karta hai ki nuclear charge (protons) kaise faila hai. Alpha particles aur neutrons, jo nuclear force mehsoos karte hain, iske bajaye nuclear matter ko map karte hain, isliye dono methods thode alag radii dete hain. Dono tarah ke experiments, bahut saare elements par, ek kamaal ke simple rule par sahmat hain:

Nuclear radius

R = R_{0} A^{1 / 3}

(13.5)

jahan

$R_{0}$: $1.2 f m = 1.2 \times 10^{- 15} m$ , har nucleus ke liye same
$A$: mass number

Cube root hi clue hai. Sphere ka volume $R^{3}$ ke saath badhta hai, isliye nucleus ka volume $R_{0}^{3} A$ ke saath badhta hai: yeh seedha nucleons ki ginti ke proportional hai. Nucleons double karne se volume double hota hai, lekin radius sirf $2^{1 / 3} = 1.26$ guna badhta hai (Fig. 13.6). Nucleus mein nucleons waise hi packed hote hain jaise jar mein marbles ya paani ki boond mein molecules: aur daalne se usi density ki badi boond banti hai. Nucleus ki yeh liquid-drop picture aage fission samajhne mein madad karegi.

2/3Unke radii 1.2 fm × 2, 3, 4 aur 5 hain: A ke cube roots.

Fig. 13.6Nucleons ko 125/8 ≈ 16 guna karo to bhi radius sirf 2.4 fm se 6.0 fm tak badhta hai: volume A ke proportional hai, isliye har nucleus ki density same hai.

Ek baseline par chaar circles khade hain, left se right bade hote hue, jin par A equals 8, 27, 64 aur 125 likha hai. Step 1 sirf circles aur unke mass numbers dikhata hai. Step 2 unke radii jodta hai, 2.4, 3.6, 4.8 aur 6.0 femtometres, yaani cube roots 2, 3, 4 aur 5 guna 1.2 fm, aur ek note ki radius A ke cube root ke saath badhta hai, saath mein formula R equals R nought A to the one-third. Step 3 har circle ko tightly packed dots ke ek hi pattern se bharta hai aur likhta hai: same packing, same density.

DerivationNuclear matter ki density

Mass number $A$ wale nucleus ka mass lagbhag $A$ atomic mass units hota hai, $m \approx A m_{u}$ jahan $m_{u} = 1.66 \times 10^{- 27} k g$ . Iska volume Eq. 13.5 se milta hai.

ρ = \frac{A m_{u}}{\frac{4}{3} π R_{0}^{3} A} = \frac{3 m_{u}}{4 π R_{0}^{3}}

Mass number cancel ho jaata hai, isliye har nucleus ki density same hai. Numbers daalne par,

ρ = \frac{3 \times 1.66 \times 10^{- 27} k g}{4 π (1.2 \times 10^{- 15} m)^{3}} \approx 2.3 \times 10^{17} k g m^{- 3}

Gold se compare karo, jo $1.9 \times 10^{4} k g m^{- 3}$ ke saath sabse dense metals mein se ek hai: nuclear matter das trillion guna zyada dense hai. Gold ki ek bar mein nuclei volume ka $10^{13}$ mein ek hisse se bhi kam bharte hain; baaki sab electrons ki patli dhundh hai.

Example 13.1NCERT Example 13.1 · p. 310

Aluminium atom ka core kitna dense hai?

Aluminium-27 atom ka mass 26.98 u hai. (a) Uske nucleus ka radius nikaalo. (b) Nucleus ki density nikaalo. (c) Aluminium metal ki density $2700 k g m^{- 3}$ hai. Nucleus kitne guna zyada dense hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Copper-64 nucleus ka radius kitna hai, femtometres mein?

Answer dekho

Answer 4.8 fm

R = 1.2 \times 64^{1 / 3} = 1.2 \times 4 = 4.8 f m

13.4 Mass-energy aur nuclear binding energy

13.4.1 Mass – Energy

1905 se pehle physics mein hisaab rakhne ke do alag rules the: reaction mein total mass same rehta hai, aur total energy bhi. Einstein ki special theory of relativity ne dono ko mila diya. Unhone dikhaya ki mass khud energy ka ek form hai, aur ise kinetic energy, light ya heat mein badla ja sakta hai, aur wapas bhi. Exchange rate vacuum mein light ki speed $c$ se fix hota hai:

Mass–energy equivalence

E = m c^{2}

(13.6)

jahan

$E$: mass ke barabar energy
$m$: mass
$c$: vacuum mein light ki speed, $3.00 \times 10^{8} m s^{- 1}$

Kyunki $c^{2}$ lagbhag $9 \times 10^{16} m^{2} s^{- 2}$ hai, isliye thoda sa mass bhi bahut zyada energy ke barabar hota hai.

Example 13.2NCERT Example 13.2 · p. 310

Fuel ke roop mein dhool ka ek kan

Maano dhool ke ek kan ka saara mass, 1.0 mg, energy mein badla ja sake. (a) Yeh kitni energy hai? (b) Yeh 1.5 kW ka air conditioner kitni der chala sakti hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Koi bhi dhool ko poori tarah energy mein nahi badal sakta, lekin nuclear reactions aur particles ka creation aur annihilation Eq. 13.6 ko seedha test karte hain, aur yeh har test mein pass hua hai. Aaj ki statement ek hi law hai: kisi bhi isolated process mein total energy conserve hoti hai, bas har mass ki energy $m c^{2}$ bhi ginni hai. Ab mass aur energy ke alag laws nahi hain, sirf conservation of mass–energy ka ek law hai. Nuclear physics iska sabse zabardast saboot deta hai, aur yeh poora chapter isi par chalta hai.

Quick check

2 mein se 1

2.0 g mass ke barabar kitni energy hai, joules mein?

Answer dekho

Answer 180000000000000 J

E = m c^{2} = (2.0 \times 10^{- 3} k g) (3.0 \times 10^{8} m s^{- 1})^{2} = 1.8 \times 10^{14} J

13.4.2 Nuclear binding energy

Nucleus protons aur neutrons se bana hai, isliye tum soch sakte ho ki uska mass uske parts ke masses ke total ke barabar hoga. Tolo to aisa nahi hota. Carbon-12 lo, jiske atom ka mass definition se exactly 12.000000 u hai. Usmein 6 protons, 6 neutrons aur 6 electrons hain. Electrons ke liye ek badhiya trick hai: protons aur electrons ko saath mein 6 hydrogen atoms gino, taaki electrons dono taraf aayein aur cancel ho jaayein. Tab parts ka weight hai

\begin{aligned} 6 m_{H} + 6 m_{n} \\ = 6.046950 u \\ + 6.051990 u \\ = 12.098940 u \end{aligned}

Jude hue atom ka mass $12.098940 - 12.000000 = 0.098940 u$ kam hai, lagbhag ek nucleon ke mass ka dasvaan hissa (Fig. 13.7). Har nucleus mein aisi kami hoti hai, jise uska mass defect (द्रव्यमान क्षति) kehte hain. Mass $M$ , $Z$ protons aur $A - Z$ neutrons wale nucleus ke liye,

Mass defect

Δ M = [Z m_{p} + (A - Z) m_{n}] - M

(13.7)

Tables se atomic masses lene par, $m_{p}$ ki jagah $m_{H}$ aur $M$ ki jagah atomic mass lo, jaise humne carbon ke liye kiya; electron masses cancel ho jaate hain. (Electrons ki apni bahut chhoti binding energies, zyada se zyada kuch keV, ignore ki jaati hain.)

3/3Nucleus bante waqt gayab mass energy ban kar nikal gaya, $Δ M c^{2} = 92.2 M e V$ .

Fig. 13.7Carbon-12 atom ke alag-alag parts khud atom se bhaari hain; gayab mass binding energy ko c² se divide karne ke barabar hai.NCERT p. 311

Stand par ek beam balance. Left pan mein baarah chhoti balls hain, chhe par H likha hai hydrogen atoms ke liye aur chhe par n neutrons ke liye, label hai 6 H atoms plus 6 n, 12.098 940 u. Right pan mein ek compact cluster hai jis par likha hai ek C-12 atom, 12.000 000 u. Step 1 balance ko barabar dikhata hai, jaise dono taraf equal hon. Step 2 beam ko jhuka deta hai taaki left pan neeche jaaye, caption ke saath: gayab mass delta M equals 0.098 940 u. Step 3 jhukaav rakhta hai aur ek highlighted card jodta hai: binding energy 92.2 MeV.

Mass kahan gaya? Mass–energy equivalence iska jawaab deti hai. Jab 6 free protons aur 6 free neutrons milkar carbon nucleus banate hain, to woh energy dete hain, aur woh energy mass $Δ M$ apne saath le jaati hai. Process ko ulta chalao: nucleus ko alag-alag nucleons mein todne ke liye tumhe utni hi energy deni padegi, aur tab alag hue parts exactly $Δ M$ zyada bhaari honge. Yeh energy nucleus ki binding energy (बंधन ऊर्जा) hai:

Binding energy

E_{b} = Δ M c^{2}

(13.8)

Binding energy hisaab rakhne wali quantity hai: experiments shayad hi kabhi nucleus ko single nucleons mein todte hain, phir bhi binding energies compare karke pata chalta hai ki kaun se nuclei zyada mazbooti se bandhe hain aur kaun si reactions energy release kar sakti hain. Eq. 13.8 ko u mein masses ke saath use karne ke liye, humein pata hona chahiye ki ek atomic mass unit kitni energy ke barabar hai.

Example 13.3NCERT Example 13.3 · p. 311

Ek atomic mass unit mein energy

(a) Electron mass ( $9.109 \times 10^{- 31} k g$ ) aur 1 u ke barabar energy nikaalo, joules aur MeV mein. $c = 2.998 \times 10^{8} m s^{- 1}$ aur $1 e V = 1.602 \times 10^{- 19} J$ lo. (b) Is result se helium-4 (atomic mass 4.002603 u) ki binding energy nikaalo, total aur per nucleon.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Carbon-12 ke liye wahi conversion $E_{b} = 0.098940 \times 931.5 M e V = 92.2 M e V$ deta hai. Bade nuclei ki binding energies naturally badi hoti hain, sirf isliye ki unmein zyada nucleons hain, isliye nuclei ko fairly compare karne ke liye hum $A$ se divide karte hain. Binding energy per nucleon,

Binding energy per nucleon

E_{b n} = \frac{E_{b}}{A}

(13.9)

binding mein har nucleon ka hissa hai: nucleus ko todne ki total keemat, uske $A$ members mein barabar baanti hui. Carbon-12 ki $92.2 / 12 = 7.68 M e V$ per nucleon hai. Har stable nucleus ke liye ise nikaalo, mass number ke against plot karo, aur ek famous curve dikhta hai (Fig. 13.8).

2/4 $A \approx 30$ se $170$ tak har nucleon lagbhag 8 se 8.8 MeV se bandha hai; peak $A = 56$ se $62$ ke paas hai.

Fig. 13.8Beech ke size wale nuclei ke liye binding energy per nucleon lagbhag constant hai aur dono ends par kam, isliye bahut bhaari nuclei ko todne aur bahut halke nuclei ko jodne, dono se energy release hoti hai.NCERT Fig. 13.1 · p. 312

Ek graph jismein mass number A, 0 se 250 tak across hai aur binding energy per nucleon 0 se 9 MeV tak upar. Lagbhag 35 stable nuclei ke blue dots ek curve follow karte hain: H-2 lagbhag 1 MeV ke paas hai, helium-4 apne halke padosiyon se upar lagbhag 7 MeV par uchhal jaata hai, carbon-12 aur oxygen-16 bhi thoda upar hain, A lagbhag 30 ke baad curve flat ho jaata hai, iron-56 par lagbhag 8.8 MeV ka peak hai, aur uranium-238 par dheere se lagbhag 7.6 MeV tak utar aata hai. Step 1 points aur trend line dikhata hai. Step 2, A 30 se 170 tak ke lagbhag flat hisse par ek dashed box banata hai, jis par likha hai: lagbhag flat, 8 se 8.8. Step 3 heavy side ko shade karta hai aur A lagbhag 236 se A lagbhag 118 tak ek arrow banata hai, jis par likha hai: fission, heavy toote. Step 4 light side ko shade karta hai aur H-2 se helium-4 tak upar ek arrow banata hai, jis par likha hai: fusion, light nuclei jude.

Curve ko left se right padho. Sabse halke nuclei ke liye $E_{b n}$ chhota hai (deuteron ke liye 1.1 MeV) aur tezi se badhta hai, aur kuch nuclei apne padosiyon se saaf upar baithe hain: helium-4 (7.07 MeV), carbon-12 aur oxygen-16. Yeh khaas nuclei is baat ka saboot hain ki nucleus ke andar nucleons shells bharte hain, bilkul waise jaise atom mein electrons shells bharte hain. Lagbhag $A = 30$ ke baad curve 8 aur 8.8 MeV per nucleon ke beech flat ho jaata hai aur $A \approx 170$ tak aisa hi rehta hai, jiska sabse ooncha point iron-56 aur nickel-62 ke aas-paas lagbhag 8.8 MeV par hai. Sabse bhaari nuclei ke liye yeh dheere-dheere neeche jhukta hai, uranium-238 ke liye 7.57 MeV tak.

Isse chaar conclusions nikalte hain, aur yahi baaki chapter ki shape tay karte hain.

Binding force bahut zyada attractive hai. Protons ke electric repulsion ke against har nucleon ko kai MeV se pakad kar rakhne ke liye ek aisa force chahiye jo ab tak dekhe kisi bhi force se bahut zyada strong ho.
Force ki range short hai. Agar har nucleon baaki har nucleon ko attract karta, to har naya nucleon baaki sabki binding mein kuch jodta, aur $E_{b n}$ , $A$ ke saath badhta hi jaata. Iske bajaye yeh flat rehta hai, jiska matlab hai ki ek nucleon sirf pahunch ke andar wale kuch padosiyon se hi interact karta hai.
Bhaari nucleus ko todne se energy release hoti hai. Uske nucleons, jo har ek lagbhag 7.6 MeV se bandhe hain, beech ke size wale nuclei mein pahunch jaate hain jahan woh lagbhag 8.4 se 8.7 MeV se bandhe hote hain. Har nucleon zyada kaske bandh jaata hai, aur farak energy ban kar nikalta hai: yahi fission hai (Section 13.7.1).
Light nuclei ko jodne se energy release hoti hai. Do deuterons (1.1 MeV per nucleon) jo milkar helium-4 (7.1 MeV per nucleon) banate hain, bahut zyada kaske bandh jaate hain. Yahi fusion hai, Sun ki energy ka source (Section 13.7.2).

Doosra conclusion dhyaan se dekhne laayak hai (Fig. 13.9). Maano pull sirf lagbhag 2 fm tak pahunchta hai, yaani lagbhag sabse paas ke padosiyon tak. Tab nucleus ke andar dabe nucleon ke charon taraf hamesha wahi chhoti bheed hoti hai, teen dimensions mein lagbhag ek darjan nucleons, chahe nucleus mein 40 nucleons hon ya 200. Door wali side par jode gaye nucleons pahunch se bahar hain aur uske liye kuch nahi badalte. Bade nucleus mein zyaadatar nucleons isi tarah andar dabe hote hain, isliye $E_{b n}$ ek constant par tik jaata hai. Sirf surface ke nucleons, jinke padosi kam hain, kam bandhe hote hain; chhote nucleus mein lagbhag saare nucleons surface par hote hain, isiliye light nuclei ke liye $E_{b n}$ kam hai. Short-range force ki is property ko, ki nucleus chahe kitna bhi bada ho, har nucleon sirf limited padosiyon se bandha rehta hai, saturation kehte hain.

1/3Nucleon par pull sirf dashed circle ke andar wale padosiyon se aata hai.

Fig. 13.9Ek nucleon sirf nuclear force ki short range ke andar wale padosiyon ko mehsoos karta hai, isliye har bade nucleus mein andar ke nucleons barabar achhe se bandhe hote hain, jabki surface ke nucleons kam bandhe hote hain.

Left mein saat grey discs ka chhota cluster; right mein saintees ka bada hexagonal cluster. Step 1: bade cluster mein beech wala nucleon purple hai, ek dashed circle pull ki range, lagbhag 2 fm, dikhata hai, aur circle ke andar uske chhe sabse paas ke padosi shaded hain. Step 2: cluster ke door wale edge par ek nucleon dashed border se outline hai aur us par likha hai ki woh bahut door hai, koi pull nahi; heading kehti hai: kisi bhi A mein same padosi. Step 3: dashed circle surface ke ek corner wale nucleon ke charon taraf bana hai, jiske andar sirf teen padosi hain; heading kehti hai: surface nucleon ke sirf 3, aur chhota cluster shaded hai aur us par likha hai: zyaadatar surface.

Quick check

4 mein se 1

Oxygen-16 ka atomic mass 15.994915 u hai.

m_{H} = 1.007825 u

m_{n} = 1.008665 u

aur

1 u = 931.5 M e V / c^{2}

use karke iski binding energy per nucleon MeV mein nikaalo.

MeV

Answer dekho

Answer 7.98 MeV

Δ M = 8 (1.007825) + 8 (1.008665) - 15.994915 = 0.137005 u

, isliye

E_{b} = 0.137005 \times 931.5 = 127.6 M e V

aur

E_{b n} = 127.6 / 16 = 7.98 M e V

13.5 Nuclear force

Atoms mein electrons Coulomb force se bandhe hote hain; bahar ke electrons, jo chemistry tay karte hain, sirf kuch eV se bandhe hain. Nucleons har ek lagbhag 8 MeV se bandhe hain, yaani lagbhag das lakh guna zyada, isliye yahan koi alag force kaam kar raha hoga. Iska kaam mushkil hai. 2 fm door do protons ek doosre ko $F = e^{2} / (4 π ε_{0} r^{2}) \approx 58 N$ ke force se repel karte hain, jo 6 kg mass ke weight jitna hai, aur yeh $10^{- 27} k g$ mass ke particle par lag raha hai. Us doori par unki electric potential energy lagbhag 0.72 MeV hai. Gravity koi madad nahi karti: unhi do protons ke beech yeh electric push se lagbhag $10^{36}$ guna kamzor hai. Jo force jeet-ta hai use nuclear force (नाभिकीय बल), ya strong force kehte hain. Zyaadatar 1930 aur 1950 ke beech hue experiments ne iske main features tay kiye.

Nuclear distances par yeh Coulomb force se bahut zyada strong hai, itna strong ki ek doosre ko repel karte protons ko kuch femtometres ke volume mein baandh de. Gravity, badle mein, dono se bahut kamzor hai.
Iski range short hai. Do nucleons ke beech yeh force tab gayab ho jaata hai jab woh kuch femtometres se zyada door hon. Saturation aur binding-energy curve ke flat hisse ki wajah yahi hai.
Moderate distances par yeh attractive hai lekin bahut chhoti distances par strongly repulsive. Do nucleons ki potential energy lagbhag 0.8 fm ke separation $r_{0}$ par sabse kam hoti hai. $r_{0}$ se zyada door hon to woh attract karte hain; $r_{0}$ se zyada paas dhakele jaayein to woh bahut strongly repel karte hain. Yahi hard core nuclei ko collapse hone se rokta hai.
Yeh charge par depend nahi karta. Do protons, do neutrons, ya ek proton aur ek neutron ke beech nuclear pull lagbhag same hota hai.

3/4 $r_{0} \approx 0.8 f m$ par potential energy sabse kam hai aur force zero hai.

Fig. 13.10Do nucleons ki schematic potential energy: door hone par koi force nahi, lagbhag 0.8 fm par minimum ki taraf slope par attraction, aur usse paas ek hard repulsive core.NCERT Fig. 13.2 · p. 313

Upar, left mein ek fixed blue nucleon aur ek orange nucleon jo ek line par chalta hai, ek dotted drop line se neeche graph par ek red dot se juda hai. Graph potential energy U (MeV mein) ko separation r (femtometres mein, 0 se 4 tak) ke against dikhata hai: blue curve door zero par flat hai, 0.8 fm par lagbhag minus 100 MeV ke minimum tak girta hai, aur lagbhag 0.6 fm se neeche tezi se upar chadh jaata hai. Step 1: nucleons 3.3 fm door hain, dot flat hisse par hai, label: door, koi force nahi. Step 2: 1.6 fm par dot minimum ki taraf jaate slope par hai, ek arrow orange nucleon ko blue ki taraf kheenchta hai, label: attraction. Step 3: 0.8 fm par dot well ke bottom par hai, label: lowest energy, 0.8 fm par. Step 4: 0.55 fm par dot steep wall par ooncha hai, ek arrow orange nucleon ko door dhakelta hai, label: strong repulsion.

Nucleon par force potential-energy curve ka negative slope hai, $F = - d U / d r$ . Jahan $r$ badhne par $U$ badhta hai ( $r_{0}$ ke baad), wahan $F$ negative hai, yaani woh chhote $r$ ki taraf point karta hai: attraction. Jahan $r$ ghatne par $U$ badhta hai ( $r_{0}$ ke andar), wahan force bahar ki taraf point karta hai: repulsion. Coulomb's law ya Newton's law of gravitation ki tarah nuclear force ka koi simple formula nahi hai, aur Fig. 13.10 ka curve sirf uski shape ka ek sketch hai.

Quick check

3 mein se 1

Do nucleons 1.5 fm door hain, jahan door jaane par unki potential energy badhti hai. Unke beech nuclear force hai

attractiverepulsivezerosirf protons ke beech attractive

Answer dekho

Answer (A) attractive

F = - d U / d r

. Agar

U

r

ke saath badhta hai, to

d U / d r > 0

aur

F < 0

: force chhote

r

ki taraf point karta hai, yaani attraction. Force charge-independent hai, isliye nucleons ke kisi bhi pair ke liye same hai.

13.6 Radioactivity

February 1896 mein Henri Becquerel test kar rahe the ki kya uranium salts, sunlight sokhne ke baad, kuch mahine pehle discover hui X-rays jaisi penetrating rays dete hain. Paris mein lagaatar baadal wale dinon ne unka experiment rok diya, isliye unhone salts ko black paper mein lipti photographic plate par ek drawer mein chhod diya. Jab unhone phir bhi plate develop ki, to woh bahut zyada fogged thi. Uranium khud hi radiation nikaal raha tha, sunlight ki koi zaroorat nahi thi. Marie aur Pierre Curie ne jaldi hi do kahin zyada active elements, polonium aur radium, dhoondhe, aur Marie Curie ne is effect ko radioactivity (रेडियोऐक्टिवता) naam diya.

Baad ke experiments ne dikhaya ki radioactivity nucleus se aati hai. Ek unstable nucleus kisi doosre nucleus mein, ya apni hi lower-energy state mein badal jaata hai, aur is process mein radiation bahar phenkta hai. Yahi radioactive decay hai. Nature mein teen tarah ke decay hote hain, jinke naam Greek alphabet ke pehle teen letters par rakhe gaye, jab kisi ko pata bhi nahi tha ki woh kya hain:

Alpha ( $α$ ) decay: nucleus ek helium-4 nucleus, $_{}^{}_{2}^{4} H e$ , emit karta hai, yaani aapas mein bandhe do protons aur do neutrons.
Beta ( $β$ ) decay: nucleus ek electron ( $β^{-}$ ) ya ek positron ( $β^{+}$ ) emit karta hai, jo electron ke mass wala lekin $+ e$ charge wala particle hai. Saath mein ek neutrino ya antineutrino bhi nikalta hai.
Gamma ( $γ$ ) decay: ek excited nucleus lower energy state mein girta hai aur high energy ka photon emit karta hai, aam taur par kuch sau keV ya usse zyada.

Table 13.1Har decay nucleus ko kaise badalta hai (har ek ka ek example)

Decay	Kya nikalta hai	Badlav	Example
$α$	$_{}^{}_{2}^{4} H e$ nucleus	Z − 2, N − 2, A − 4	$_{}^{}_{88}^{226} R a \overset{}{\to}_{}^{}_{86}^{222} R n +_{}^{}_{2}^{4} H e$
$β^{-}$	electron, antineutrino	Z + 1, N − 1	$_{}^{}_{6}^{14} C \overset{}{\to}_{}^{}_{7}^{14} N + e^{-} + \overset{ˉ}{ν}$
$β^{+}$	positron, neutrino	Z − 1, N + 1	$_{}^{}_{11}^{22} N a \overset{}{\to}_{}^{}_{10}^{22} N e + e^{+} + ν$
$γ$	photon	koi nahi	$_{}^{}_{28}^{60} N i^{*} \overset{}{\to}_{}^{}_{28}^{60} N i + γ$

Har decay mein charge aur nucleons ki total ginti conserve hoti hai: Table 13.1 ke examples mein har side ke upar wale numbers aur neeche wale numbers jodo, woh match karte hain. Nucleus ke andar $β^{-}$ decay ek neutron ko proton mein badalta hai, aur $β^{+}$ decay ek proton ko neutron mein. Nuclide chart par har decay ek fixed step hai (Fig. 13.11).

2/5Alpha decay 2 protons aur 2 neutrons hata deta hai: do step neeche aur do left.

Fig. 13.11Har tarah ka decay nucleus ko N–Z chart par ek fixed step se khiskata hai; gamma decay use usi cell mein chhodta hai.

Cells ka ek grid jismein N right ki taraf aur Z upar ki taraf badhta hai; parent nucleus, jis par Z, N likha hai, beech wale cell mein hai. Step 1 sirf parent dikhata hai. Step 2: ek arrow do cell left aur do neeche ek dashed daughter cell tak jaata hai, aur side panel par likha hai Z minus 2, N minus 2, A minus 4: alpha decay. Step 3: ek arrow ek cell left aur ek upar jaata hai, likha hai Z plus 1, N minus 1, A same: beta-minus decay. Step 4: ek arrow ek cell right aur ek neeche jaata hai, likha hai Z minus 1, N plus 1, A same: beta-plus decay. Step 5: ek ghoomta arrow usi cell mein wapas aata hai, likha hai Z, N, A same hain; energy girti hai: gamma decay.

Kuch nuclei unstable kyun hote hain? Light stable nuclei mein lagbhag jitne protons utne hi neutrons hote hain ( $N / Z \approx 1$ ). Bhaari nuclei ko extra neutrons chahiye: unke bahut saare protons poore nucleus mein ek doosre ko repel karte hain, jabki nuclear attraction sirf sabse paas ke padosiyon tak pahunchta hai, aur extra neutrons repulsion badhaye bina attraction jodte hain. Lead-208 tak ratio $126 / 82 \approx 1.5$ ho jaata hai. Apne size ke hisaab se zyada neutrons wala nucleus aksar $β^{-}$ decay karta hai, zyada protons wala $β^{+}$ decay, aur bahut bhaari nuclei aksar ek alpha particle nikaal dete hain. Stability exception hai: 3000 se zyada jaane-maane nuclides mein se das mein ek se bhi kam stable hai. Baaki labs mein stable nuclei par protons, neutrons, deuterons ya alpha particles bombard karke banaye jaate hain, ya stars ke malbe mein dikhte hain.

Quick check

3 mein se 1

Uranium-238 (

_{}^{}_{92}^{238} U

) alpha decay karta hai. Daughter nucleus hai

_{}^{}_{90}^{234} T h

_{}^{}_{90}^{238} T h

_{}^{}_{92}^{234} U

_{}^{}_{91}^{236} P a

Answer dekho

Answer (A) $_{}^{}_{90}^{234} T h$

Alpha particle 2 protons aur 2 neutrons le jaata hai:

Z = 92 - 2 = 90

(thorium),

A = 238 - 4 = 234

A = 238

rakhne ka matlab hai bhool jaana ki alpha particle 4 nucleons le jaata hai.

13.7 Nuclear energy

Binding-energy curve yeh bhi map karta hai ki energy kahan mil sakti hai. Iska beech wala hissa lagbhag 8 MeV per nucleon par ek lamba plateau hai, aur dono ends neeche hain. Zyada kaske bandhe system ka mass kam hota hai, isliye jab bhi nucleons dheele bandhe nuclei se zyada kaske bandhe nuclei mein rearrange hote hain, mass gayab hota hai aur energy nikalti hai. Plateau ki taraf chadhne ke do raaste hain: ek bahut bhaari nucleus ko do beech ke size wale nuclei mein todo (fission), ya do bahut halke nuclei ko jodkar ek bhaari nucleus banao (fusion).

Kisi nuclear process mein nikli energy uski Q-value hai. Conservation of mass–energy se ise kinetic energies se ya masses se, dono tarah nikaala ja sakta hai:

Q-value

\begin{aligned} Q & = K_{final} - K_{initial} \\ = (\sum m_{initial} - \sum m_{final}) c^{2} \end{aligned}

Positive $Q$ ka matlab hai energy release hoti hai (process exothermic hai); negative $Q$ ka matlab hai energy deni padti hai (endothermic). Kyunki protons aur neutrons sirf regroup hote hain, $Q$ baad ki total binding energy minus pehle ki total binding energy ke barabar bhi hai.

Ise chemistry se compare karo. Coal, petrol aur cooking gas atoms ke beech electrons ko rearrange karke energy dete hain, per atom kuch eV: ek carbon atom ko jalakar carbon dioxide banane se lagbhag 4.1 eV milta hai. Ek single uranium fission lagbhag 200 MeV deta hai, yaani lagbhag paanch crore guna zyada. Per kilogram, uranium-235 ka complete fission lagbhag $8 \times 10^{13} J$ dega, jabki ek kilogram achha coal lagbhag $3 \times 10^{7} J$ deta hai. Kilogram ke hisaab se nuclear fuel chemical fuel se kuch million guna zyada amir hai. Neeche wale lab mein curve aur dono tarah ki reactions explore karo.

InteractiveBinding-energy curve lab

Lagbhag 35 stable nuclides ke liye binding energy per nucleon ka mass number ke against graph, teen modes ke saath. Read mode mein ek slider nuclide chunta hai aur cards uski binding energy per nucleon, total binding energy aur mass number dikhate hain. Split mode mein ek slider uranium-236 nucleus se bane halke fragment ka mass number set karta hai aur ek menu do ya teen free neutrons set karta hai; markers aur dashed lines curve par parent aur dono fragments dikhate hain, aur cards pehle aur baad ki total binding energy aur release hui energy dete hain, lagbhag 160 se 205 MeV. Join mode mein ek menu saat fusion reactions mein se ek chunta hai, jaise deuterium plus tritium, aur cards pehle aur baad ki binding energies aur Q-value dete hain.

Fig. 13.12Curve se binding energies padho, uranium-236 nucleus ko apni pasand ke fragments mein todo, ya koi fusion reaction chuno, aur dekho kitni energy release hoti hai.

13.7.1 Fission

Natural radioactivity dheemi hai aur ise on ya off nahi kiya ja sakta. Kahin zyada control nuclear reactions se milta hai, jinmein hum nuclei par protons, deuterons, alpha particles ya neutrons jaise particles chalate hain. Neutrons sabse achhi goliyaan hain: uncharged hone ki wajah se woh koi Coulomb repulsion mehsoos nahi karte aur dheere chalte hue bhi nucleus tak pahunch sakte hain. 1938 mein Otto Hahn aur Fritz Strassmann ne uranium par neutrons bombard kiye aur hairaani se products mein barium paaya, jo lagbhag aadha bhaari element hai. Lise Meitner aur Otto Frisch ne kuch hi hafton mein ise explain kar diya: uranium nucleus do hisson mein toot gaya tha. Unhone ise nuclear fission (नाभिकीय विखंडन) kaha.

Jab uranium-235 ek slow neutron absorb karta hai, to woh bahut excited state mein uranium-236 ban jaata hai, jise $_{}^{}_{92}^{236} U^{*}$ likhte hain. Yeh nucleus chot khaayi paani ki boond ki tarah hilta hai aur zyaadatar cases mein do medium-sized fragments aur kuch neutrons mein toot jaata hai (Fig. 13.13):

Neutron capture

_{}^{}_{0}^{1} n +_{}^{}_{92}^{235} U \overset{}{\to}_{}^{}_{92}^{236} U^{*}

Split har baar same nahi hota. Excited nucleus ke tootne ke bahut se tareekon mein se teen yeh hain

\begin{aligned} _{}^{}_{}^{236} U^{*} & ⟶_{}^{}_{}^{141} B a \\ +_{}^{}_{}^{92} K r + 3 n \end{aligned}

(13.10)

\begin{aligned} _{}^{}_{}^{236} U^{*} & ⟶_{}^{}_{}^{139} X e \\ +_{}^{}_{}^{95} S r + 2 n \end{aligned}

(13.11)

\begin{aligned} _{}^{}_{}^{236} U^{*} & ⟶_{}^{}_{}^{140} C s \\ +_{}^{}_{}^{93} R b + 3 n \end{aligned}

(13.12)

Har case mein charges bhi balance hote hain: barium, krypton, xenon, strontium, caesium aur rubidium ke $Z = 56, 36, 54, 38, 55$ aur $37$ hain, aur har pair jodkar uranium ka 92 deta hai.

4/4Do fragments 2 ya 3 neutrons ke saath door udte hain; lagbhag 200 MeV release hota hai.

Fig. 13.13Liquid-drop picture mein, excited compound nucleus tab tak khinchta hai jab tak uske do halves ke beech electric repulsion neck ke across nuclear attraction par jeet nahi jaata.

Step 1: n mark ki hui ek chhoti blue ball, jis par slow likha hai, U-235 likhe ek bade orange circle ki taraf badhti hai. Step 2: neutron absorb ho gaya hai aur drop side mein khinchkar oval ban gaya hai, label: U-236 star, excited aur hilta hua. Step 3: drop ke beech mein moongphali jaisi patli kamar hai, label: neck banti hai. Step 4: do alag size ke circles, har ek par fragment likha hai, left aur right door ud gaye hain, aur teen chhote neutrons bahar udte hain, label: 2 fragments plus 2 ya 3 neutrons plus lagbhag 200 MeV.

Liquid-drop picture batati hai ki split kyun hota hai. Jaise-jaise excited drop khinchti hai, nuclear attraction, jo sirf padosiyon ke beech lagta hai, surface ko kam achhe se pakad paata hai, jabki uske do ends ke beech electric repulsion, jo poore nucleus mein lagta hai, strong bana rehta hai. Ek had se zyada khinchne par repulsion jeet jaata hai. Fragments lagbhag uranium ke neutron-to-proton ratio, lagbhag 1.55, ke saath bante hain, jo medium nuclei ke liye bahut zyada neutron-rich hai. Isliye woh radioactive hote hain aur $β^{-}$ decays ki ek chain se stability tak pahunchte hain; isiliye reactor ka spent fuel itna khatarnaak hota hai.

Example

Ek fission ki energy ka estimate

Binding-energy curve use karke us energy ka estimate lagao jo tab release hoti hai jab $_{}^{}_{}^{236} U^{*}$ , $A = 95$ aur $A = 139$ wale fragments aur do free neutrons mein toot-ta hai. $A = 236$ ke liye $E_{b n} \approx 7.59 M e V$ , $A = 95$ ke liye $8.66 M e V$ aur $A = 139$ ke liye $8.38 M e V$ padho.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Ismein se lagbhag 170 MeV shuru mein do fragments le jaate hain, jo apne aapas ke electric repulsion se door phenke jaate hain. Baaki neutrons, gamma rays, aur fragments ke baad ke beta decays mein jaata hai, jismein bahar nikal jaane wale antineutrinos bhi shamil hain. Fragments aas-paas ke fuel mein ek millimetre ke chhote se hisse mein ruk jaate hain, isliye unki energy heat ban jaati hai. Yahi heat har nuclear power station ka energy source hai: yeh paani ubaalti hai, aur steam turbines ko bilkul coal plant ki tarah ghumati hai. Kyunki har fission 2 ya 3 neutrons bhi release karta hai, ek fission doosre fissions ko trigger kar sakta hai. Reactor mein yeh chain har fission par ek naye fission par steady rakhi jaati hai; nuclear weapon mein ise bekaabu hone diya jaata hai, jo second ke chhote se hisse mein bahut zyada energy release karta hai.

Quick check

3 mein se 1

Uranium mein fission trigger karne ke liye protons ke bajaye neutrons kyun use hote hain?

Neutrons protons se bahut halke hote hainNeutrons nucleus se koi Coulomb repulsion mehsoos nahi karteRoom temperature par neutrons ke paas zyada kinetic energy hoti haiProtons ko koi bhi nucleus absorb nahi kar sakta

Answer dekho

Answer (B) Neutrons nucleus se koi Coulomb repulsion mehsoos nahi karte

Uranium nucleus (

Z = 92

) ki taraf aate proton ko 10 MeV se zyada ka Coulomb barrier chadhna padta hai; uncharged neutron, chahe slow ho, seedha andar chala jaata hai. Asal mein neutrons protons se thoda bhaari hote hain.

13.7.2 Nuclear fusion – stars mein energy generation

Binding-energy curve ke left mein energy nuclei ko jodne se aati hai. Jab do light nuclei milkar ek bhaari, zyada kaske bandha nucleus banate hain, to yeh process nuclear fusion (नाभिकीय संलयन) hai. Kuch important fusion reactions, atomic masses se nikaale gaye Q-values ke saath, Table 13.2 mein diye gaye hain.

Table 13.2Kuch fusion reactions aur unke Q-valuesNCERT Eq. 13.13 · p. 315

Reaction	Q (MeV)	Kahan kaam aata hai
$_{}^{}_{}^{1} H +_{}^{}_{}^{1} H \overset{}{\to}_{}^{}_{}^{2} H + e^{+} + ν$	0.42	Sun mein pehla, sabse dheema step
$_{}^{}_{}^{2} H +_{}^{}_{}^{2} H \overset{}{\to}_{}^{}_{}^{3} H e + n$	3.27	deuterium fusion (ek branch)
$_{}^{}_{}^{2} H +_{}^{}_{}^{2} H \overset{}{\to}_{}^{}_{}^{3} H +_{}^{}_{}^{1} H$	4.03	deuterium fusion (doosri branch)
$_{}^{}_{}^{2} H +_{}^{}_{}^{3} H \overset{}{\to}_{}^{}_{}^{4} H e + n$	17.59	experimental fusion reactors ka fuel

Fusion mein ek pench hai. Short-range nuclear force ko dono nuclei pakadne ke liye, unhe ek doosre se kuch femtometres ke andar aana hoga, lekin dono positively charged hain aur repel karte hain. Isliye unhe itni kinetic energy ke saath aana hoga ki woh Coulomb barrier par chadh sakein (Fig. 13.14). Barrier ki height dono nuclei ke charges ke saath badhti hai, isliye yeh hydrogen isotopes ke liye sabse kam, kuch sau keV, hoti hai.

1/4Contact ke bahar electric repulsion ek pahaadi banata hai; andar nuclear force ek gehra well banata hai.

Fig. 13.14Nuclear force unhe baandh sake, isse pehle do nuclei ko apna electric repulsion paar karna hota hai; stars mein kuch unusually tez nuclei quantum tunnelling ki madad se aisa kar paate hain.NCERT Points to ponder 5 · p. 316

Separation r ke against potential energy ka ek blue curve. Right se aate hue nuclei ke paas aane par curve dheere-dheere upar uthta hai, ek Coulomb pahaadi jiske top par, contact distance par, likha hai: D–T ke liye lagbhag 0.44 MeV; thoda andar yeh seedha neeche ek gehre nuclear well mein girta hai. Step 1 sirf curve dikhata hai. Step 2: ek neechi dashed red energy line right se aati hai aur pahaadi par kaafi bahar ruk jaati hai, arrows nucleus ko wapas mudta dikhate hain, label: bahut slow, wapas mudta hai. Step 3: pahaadi ke top se upar ek dashed energy line poori andar tak jaati hai, label: kaafi tez, top ke upar se. Step 4: ek medium energy line pahaadi tak pahunchti hai aur ek dotted arrow seedha usmein se hokar well mein jaata hai, label: koi rare tez nucleus tunnel karta hai.

Example

Fusion fuel kitna garam hona chahiye?

Ek deuteron aur ek triton (tritium ka nucleus) chhoote hi fuse ho jaate hain. Unhe $R = 1.2 A^{1 / 3} f m$ wale spheres maano. (a) Coulomb barrier ki height nikaalo. (b) Kis temperature par nucleus ki average kinetic energy, $\frac{3}{2} k_{B} T$ , is barrier ke barabar hogi? ( $k_{B} = 1.38 \times 10^{- 23} J K^{- 1}$ .)

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Matter ko itna garam karke ki uske nuclei barrier paar karne laayak tez chalne lagein, jo fusion hota hai use thermonuclear fusion kehte hain. Yahi har star ko power deta hai. Phir bhi Sun ke centre ka temperature lagbhag $1.5 \times 10^{7} K$ hai, jahan proton ki average kinetic energy, $\frac{3}{2} k_{B} T$ , sirf lagbhag 2 keV hai. Do cheezein Sun ko bachaati hain. Garam gas mein speeds faili hui hoti hain, isliye protons ka ek bahut chhota hissa average se kai guna tez chalta hai; aur quantum mechanics aise proton ko barrier ke upar chadhe bina uske top se tunnel karne deti hai. Core mein har proton ko fuse hone ke liye average arabon saal intezaar karna padta hai, aur isi dheemepan ki wajah se Sun itne lambe samay tak steadily chamakta hai.

Sun hydrogen ko proton–proton (p–p) chain se helium mein badalta hai (Fig. 13.15, Table 13.3):

Table 13.3p–p chain ke stepsNCERT Eq. 13.14 · p. 316

Step	Reaction	Q (MeV)
(i)	$_{}^{}_{}^{1} H +_{}^{}_{}^{1} H \overset{}{\to}_{}^{}_{}^{2} H + e^{+} + ν$	0.42
(ii)	$e^{+} + e^{-} \overset{}{\to} 2 γ$	1.02
(iii)	$_{}^{}_{}^{2} H +_{}^{}_{}^{1} H \overset{}{\to}_{}^{}_{}^{3} H e + γ$	5.49
(iv)	$_{}^{}_{}^{3} H e +_{}^{}_{}^{3} H e \overset{}{\to}_{}^{}_{}^{4} H e + 2_{}^{}_{}^{1} H$	12.86

Yahan $_{}^{}_{}^{1} H$ ka matlab proton aur $_{}^{}_{}^{2} H$ ka matlab deuteron hai. Step (ii) mein, step (i) mein bana har positron plasma ke kisi free electron se milta hai aur pair annihilate ho jaata hai.

Reaction (iv) do helium-3 nuclei khapata hai, isliye uske har ek baar chalne ke liye pehli teen reactions ko do baar chalna padta hai. (i)–(iii) ko double karke aur (iv) jodkar, (iv) mein wapas mile do protons andar gaye chhe mein se do ko cancel kar dete hain, aur hisaab yun balance hota hai

p–p chain ka net result

\begin{aligned} 4_{}^{}_{}^{1} H + 2 e^{-} \\ ⟶_{}^{}_{}^{4} H e + 2 ν + 6 γ \\ + 26.7 M e V \end{aligned}

(13.15)

Dono taraf do aur electrons jodne se yeh poore atoms ki statement ban jaati hai: chaar hydrogen atoms ek helium atom bante hain aur 26.7 MeV release karte hain. Atomic masses se check karo: $4 (1.007825) - 4.002603 = 0.028697 u$ , aur $0.028697 \times 931.5 M e V = 26.7 M e V$ .

4/4Steps (i)–(iii) do baar hote hain, aur har positron ek electron ke saath annihilate hota hai (step ii).

Fig. 13.15p–p chain mein chaar protons aakhir mein ek helium-4 nucleus bante hain; har step aur do positron annihilations ko ginne par 26.7 MeV milta hai.

Plus mark wali balls protons hain aur n mark wali balls neutrons. Step 1, upar ki row: do single protons milkar deuteron banate hain, ek proton aur ek neutron, saath mein ek positron aur ek neutrino, 0.42 MeV. Step 2 doosri row jodta hai: ek deuteron aur ek proton helium-3 banate hain, do protons aur ek neutron, saath mein ek gamma ray, 5.49 MeV. Step 3 teesri row jodta hai aur pehli do ko dim karta hai: do helium-3 nuclei helium-4 banate hain, do protons aur do neutrons, aur do protons release karte hain, 12.86 MeV. Step 4 pehli do rows par times two mark karta hai aur tally ke saath ek highlighted card jodta hai: 0.42, 1.02 aur 5.49 ke sum ka do guna, plus 12.86, barabar 26.7 MeV.

Example

Sun apna fuel kitni tezi se use karta hai?

Sun $3.8 \times 10^{26} W$ radiate karta hai. (a) Woh har second kitna mass khota hai? (b) Agar har 26.7 MeV chaar hydrogen atoms (atomic mass 1.0078 u) se aata hai, to woh har second kitna hydrogen helium mein badalta hai? Neutrinos dwara le jaayi gayi energy ke chhote hisse ko ignore karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Sun lagbhag 4.6 billion saal purana hai aur uske core mein itna hydrogen hai ki woh lagbhag aur 5 billion saal chal sake. Jab core ka hydrogen kam padne lagta hai, to wahan fusion dheema hota hai, aur core gravity se sikud kar garam hota hai. Lagbhag $10^{8} K$ par helium nuclei khud fuse hone lagte hain: teen helium-4 nuclei ek carbon-12 banate hain aur 7.3 MeV release karte hain. Isi beech bahar ki layers bahut zyada phool jaati hain aur Sun ek red giant ban jaata hai. Sun se kahin bhaari stars mein core aur bhi garam hota hai aur oxygen, neon, silicon aur aakhir mein iron banata hai. Wahan chain ruk jaati hai, kyunki iron binding-energy curve ke top par hai: fusion se usse bhaari kuch bhi banana energy absorb karta hai. Iron se aage ke elements, jaise jewellery ka gold aur reactors ka uranium, explode hote stars mein aur neutron stars ki takkar mein neutron capture se bane.

Quick check

3 mein se 1

Fusion ko millions of kelvin ka temperature kyun chahiye, jabki fission slow neutrons se shuru ho sakta hai?

Fusion reactions fission se kam energy release karti hainLight nuclei heavy nuclei se zyada kaske bandhe hote hainFusion ke liye neutrons chahiye jo sirf high temperature par bante hainDo positive nuclei ko paas aane ke liye apna electric repulsion paar karna padta hai

Answer dekho

Answer (D) Do positive nuclei ko paas aane ke liye apna electric repulsion paar karna padta hai

Fuse hone wale dono partners positively charged hain, isliye unhe Coulomb barrier paar karne laayak kinetic energy ke saath aana padta hai; heat yahi energy deti hai. Fission trigger karne wale neutron par koi charge nahi hota aur uske saamne koi barrier nahi hota.

13.7.3 Controlled thermonuclear fusion

Kya hum Earth par ek star bana sakte hain? Fusion power plant deuterium jalayega, jo samudra ke paani mein bahut hai, tritium ke saath, jise lithium se banaya ja sakta hai, aur koi long-lived fission fragments nahi chhodega. Fuel ko lagbhag $10^{8} K$ tak garam karna padta hai, jo Sun ke core se kai guna zyada garam hai, kyunki reactor rare tunnelling events ke liye arabon saal intezaar nahi kar sakta. Itne temperature par atoms ke electrons alag ho jaate hain, aur fuel ek plasma ban jaata hai, yaani ions aur free electrons ki gas. Challenge ise pakad kar rakhna hai: koi bhi solid wall plasma ko turant thanda kar degi aur khud bhi barbaad ho jaayegi. Tokamak mein, doughnut ki shape wale strong magnetic fields charged particles ko walls se door ghumaate rehte hain (Part I ke Chapter 4 se yaad karo ki magnetic field chalte charges ko kaise modta hai). Ek aur tareeka powerful laser beams se fuel ke chhote pellets ko kuchalta hai.

Example 13.4NCERT Example 13.4 · p. 317

Check karne ke liye teen claims

Har claim ko test karo. (a) Eq. 13.10 balance nahi ho sakti, kyunki uranium andar jaata hai aur barium aur krypton bahar aate hain. Ise $C H_{4} + 2 O_{2} \overset{}{\to} C O_{2} + 2 H_{2} O$ jaisi chemical equation se compare karo. (b) Agar reaction mein har proton aur neutron bacha rehta hai, to koi mass gayab nahi ho sakta. (c) Mass sirf nuclear reactions mein energy banta hai, fuel jalne par kabhi nahi. 12 g carbon jalane se $3.94 \times 10^{5} J$ release hota hai; energy bane mass ke fraction ko uranium-235 ke ek fission (200 MeV, $A = 236$ ) wale fraction se compare karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

To claim (a) galat hai, lekin yeh ek asli farak ki taraf ishara karta hai. Chemical equation har element ke atoms ko balance karti hai: carbon, hydrogen aur oxygen atoms sirf regroup hote hain. Nuclear reaction mein elements khud badal jaate hain (uranium barium aur krypton ban jaata hai), lekin is jaisi reactions mein protons ki ginti aur neutrons ki ginti alag-alag conserve hoti hai; beta decay, jo ek ko doosre mein badalta hai, exception hai. (Bahut high energies par yeh bhi fail ho sakta hai; jo hamesha conserve hota hai woh hai total charge aur nucleons aur related particles ki total ginti, yaani baryon number.)

Quick check

2 mein se 1

Fusion reactor ka plasma material walls ke bajaye magnetic fields se mainly isliye pakda jaata hai kyunki

chhoone wali koi bhi wall plasma ko thanda kar degi aur khud damage hogimagnetic fields plasma ko

10^{8} K

tak garam karte hainnuclei neutral hote hain aur walls ke paar nikal jaate hainwalls saare neutrons absorb kar lengi

Answer dekho

Answer (A) chhoone wali koi bhi wall plasma ko thanda kar degi aur khud damage hogi

Wall se contact plasma ko turant fusion temperature se neeche thanda kar dega (aur wall ghis jaayegi). Magnetic fields charged ions aur electrons par unhe chhooye bina kaam karte hain. Plasma mein nuclei charged hote hain, neutral nahi.

Recap

Summary aur key points

Nucleus radius mein atom se lagbhag $10^{4}$ guna chhota hai lekin uske mass ka 99.9 % se zyada rakhta hai (Section 13.1).
Atomic masses atomic mass units mein naape jaate hain: $1 u$ carbon-12 atom ke mass ka barahvaan hissa hai, $1.6605 \times 10^{- 27} k g$ (Section 13.2).
Nucleus mein $Z$ protons aur $N$ neutrons hote hain; mass number $A = Z + N$ hai aur nuclide ko $_{}^{}_{Z}^{A} X$ likhte hain. Neutron, jise Chadwick ne 1932 mein dhoondha, proton se thoda bhaari hai (Neutron ki discovery).
Isotopes ka $Z$ same hota hai, isobars ka $A$ , isotones ka $N$ . Kisi element ka atomic mass uske isotope masses ka abundance-weighted average hai (Section 13.2).
Scattering experiments $R = R_{0} A^{1 / 3}$ dete hain jahan $R_{0} = 1.2 f m$ , isliye sabhi nuclei ki density same hai, lagbhag $2.3 \times 10^{17} k g m^{- 3}$ (Section 13.3).
Mass energy ka ek form hai, $E = m c^{2}$ ; total energy, $m c^{2}$ ko milakar, conserve hoti hai (Section 13.4.1).
Nucleus apne alag-alag nucleons se mass defect $Δ M = [Z m_{p} + (A - Z) m_{n}] - M$ jitna halka hai; uski binding energy $E_{b} = Δ M c^{2}$ hai, jahan $1 u \equiv 931.5 M e V$ (Section 13.4.2).
$30 < A < 170$ ke liye binding energy per nucleon lagbhag 8 MeV hai, iron ke aas-paas lagbhag 8.8 MeV par peak hai aur light aur bahut heavy nuclei ke liye kam hai (Section 13.4.2).
Nuclear force strong, short-ranged (kuch fm), lagbhag 0.8 fm se neeche repulsive aur charge se independent hai; iski short range saturation ko explain karti hai (Section 13.5).
Radioactive nuclei alpha particles, electrons ya positrons, ya gamma rays emit karte hain; har decay mein charge aur nucleon number conserve hote hain (Section 13.6).
Kisi process ki Q-value $(\sum m_{initial} - \sum m_{final}) c^{2}$ hai. Nuclear energies per event MeV mein hoti hain jabki chemistry mein eV, yaani das lakh guna ka farak (Section 13.7).
Uranium-235 jaise bhaari nucleus ka fission lagbhag 200 MeV release karta hai; light nuclei ka fusion stars ko power deta hai, jismein p–p chain chaar hydrogen atoms ko helium aur 26.7 MeV mein badalti hai, aur Coulomb barrier paar karne ke liye bahut high temperature chahiye (Section 13.7.1, Section 13.7.2, Section 13.7.3).

Table 13.4Is chapter ki quantitiesNCERT p. 319

Quantity	Symbol	Unit	Typical size
Atomic mass unit	u	$1 u = 1.6605 \times 10^{- 27} k g$	proton 1.0073 u
Nuclear radius	$R = R_{0} A^{1 / 3}$	fm ( $10^{- 15} m$ )	1.2 fm (A = 1) se 7.4 fm (A = 238) tak
Mass defect	$Δ M$	u	0.030 u (helium-4) se 1.93 u (uranium-238) tak
Binding energy	$E_{b}$	MeV	2.2 MeV (deuteron) se 1800 MeV (uranium-238) tak
Binding energy per nucleon	$E_{b n}$	MeV	1.1 MeV (deuteron) se 8.8 MeV (iron, nickel) tak
Q-value	$Q$	MeV	0.42 MeV (p + p) se lagbhag 200 MeV (fission) tak
Decay constant	$λ$	$s^{- 1}$	per second decay ka chance
Half-life	$T_{1 / 2}$	s	carbon-14 ke liye 5730 saal
Mean life	$τ$	s	free neutron ke liye lagbhag 880 s
Activity	$R$	Bq (per second decays)	nuclei ki ginti aur $λ$ ke saath badhti hai

Flash cards · 10

Formulas ka revision

1 / 10

Practice · 16 questions

Practice questions answers ke saath (NCERT “Exercises”)

NumericalEasyNCERT Exercises 13.1
Neon-20 ka atomic mass 19.992440 u hai. Iski binding energy MeV mein nikaalo. Jahan zaroorat ho, yeh use karo: $m_{H} = 1.007825 u$ , $m_{n} = 1.008665 u$ , $1 u = 931.5 M e V / c^{2}$ , $1 M e V = 1.602 \times 10^{- 13} J$ , $N_{A} = 6.022 \times 10^{23} m o l^{- 1}$ , $e^{2} / 4 π ε_{0} = 1.44 M e V f m$ , $R_{0} = 1.2 f m$ , 1 saal $= 3.156 \times 10^{7} s$ .
MeV
Solution dikhao
Neon-20 mein 10 protons aur 10 neutrons hain. Atomic masses use karne par electrons cancel ho jaate hain:
$Δ M = 10 (1.007825) + 10 (1.008665) - 19.992440 = 0.172460 u$
$E_{b} = 0.172460 \times 931.5 M e V = 160.6 M e V$
Yeh 8.03 MeV per nucleon hai.
2
NumericalMediumNCERT Exercises 13.2
Calcium-40 (atomic mass 39.962591 u) aur lead-208 (atomic mass 207.976652 u) dono bahut stable nuclei hain.
1. Calcium-40 ki binding energy MeV mein nikaalo.
  MeV
  Solution dikhao
  $Δ M = 20 (1.007825) + 20 (1.008665) - 39.962591 = 0.367209 u$
  $E_{b} = 0.367209 \times 931.5 = 342.1 M e V$
2. Lead-208 ki binding energy MeV mein nikaalo.
  MeV
  Solution dikhao
  $Δ M = 82 (1.007825) + 126 (1.008665) - 207.976652 = 1.756788 u$
  $E_{b} = 1.756788 \times 931.5 = 1636 M e V$
3. Kaun sa nucleus apne nucleons ko zyada kaske pakadta hai?
  calcium-40lead-208dono barabar
  Solution dikhao
  Per nucleon: calcium-40, $342.1 / 40 = 8.55 M e V$ ; lead-208, $1636 / 208 = 7.87 M e V$ . Calcium-40 zyada kaske bandha hai, halaanki lead-208 ki total binding energy zyada hai.
NumericalMediumNCERT Exercises 13.3
Aluminium foil ki 5.0 g ki ek strip pure aluminium-27 (atomic mass 26.981539 u) hai. Usmein har proton aur neutron ko baaki sabse alag karne ke liye kitni energy chahiye hogi? Ise 1-kiloton explosion ki energy, $4.2 \times 10^{12} J$ , se compare karo.
J
Solution dikhao
Ek aluminium-27 nucleus (13 protons, 14 neutrons) ki binding energy:
$Δ M = 13 (1.007825) + 14 (1.008665) - 26.981539 = 0.241496 u, E_{b} = 0.241496 \times 931.5 = 225.0 M e V$
5.0 g mein atoms ki ginti:
$N = \frac{5.0 g}{26.98 g m o l^{- 1}} \times 6.022 \times 10^{23} m o l^{- 1} = 1.116 \times 10^{23}$
$E = N E_{b} = 1.116 \times 10^{23} \times 225.0 M e V = 2.51 \times 10^{25} M e V = 4.0 \times 10^{12} J$
Yeh lagbhag 1-kiloton explosion jitni energy hai, jo sirf kuch grams foil ke nuclei ko todne ke liye chahiye.
NumericalEasyNCERT Exercises 13.4
Uranium-238 nucleus aur zinc-64 nucleus ke radius ka ratio estimate karo. Unke volumes ka ratio kya hai?
Solution dikhao
$\frac{R_{U}}{R_{Z n}} = {(\frac{238}{64})}^{1 / 3} = (3.72)^{1 / 3} = 1.55$
Volumes mass numbers ke ratio mein hain, $238 / 64 = 3.72$ , kyunki volume $A$ ke proportional hai.
5
NumericalMediumNCERT Exercises 13.5
Neeche do historic reactions di gayi hain. $Q = (total mass before - total mass after) c^{2}$ aur atomic masses $_{}^{}_{}^{1} H$ 1.007825 u, $_{}^{}_{}^{4} H e$ 4.002603 u, $_{}^{}_{}^{7} L i$ 7.016003 u, $_{}^{}_{}^{14} N$ 14.003074 u aur $_{}^{}_{}^{17} O$ 16.999132 u use karke, neeche har historic reaction ke liye $Q$ nikaalo aur batao ki woh energy release karti hai ya absorb.
1. $_{}^{}_{3}^{7} L i +_{}^{}_{1}^{1} H \overset{}{\to}_{}^{}_{2}^{4} H e +_{}^{}_{2}^{4} H e$ (accelerator se toda gaya pehla nucleus, Cockcroft aur Walton, 1932). $Q$ MeV mein batao.
  MeV
  Solution dikhao
  $Δ m = 7.016003 + 1.007825 - 2 (4.002603) = 0.018622 u$
  $Q = 0.018622 \times 931.5 = + 17.35 M e V$
  $Q > 0$ : exothermic. Electron masses balance ho jaate hain kyunki dono taraf atomic numbers ka total same hai ( $3 + 1 = 2 + 2$ ).
2. $_{}^{}_{7}^{14} N +_{}^{}_{2}^{4} H e \overset{}{\to}_{}^{}_{8}^{17} O +_{}^{}_{1}^{1} H$ (pehla artificial transmutation, Rutherford, 1919). $Q$ MeV mein batao.
  MeV
  Solution dikhao
  $Δ m = 14.003074 + 4.002603 - 16.999132 - 1.007825 = - 0.001280 u$
  $Q = - 0.001280 \times 931.5 = - 1.19 M e V$
  $Q < 0$ : endothermic. Alpha particle ko kam se kam itni kinetic energy laani padegi (asal mein zyada, kyunki products ko momentum bhi le jaana hota hai).
NumericalMediumNCERT Exercises 13.6
Kya calcium-40 nucleus (atomic mass 39.962591 u) do neon-20 nuclei (har ek ka atomic mass 19.992440 u) mein toot kar energy release kar sakta hai? $Q$ nikaal kar tay karo.
MeV
Solution dikhao
$Q = [39.962591 - 2 (19.992440)] \times 931.5 M e V = (- 0.022289) (931.5) M e V = - 20.8 M e V$
$Q$ negative hai: dono neon nuclei milkar calcium nucleus se bhaari hain. Split ke liye lagbhag 21 MeV energy deni padegi, isliye yeh apne aap nahi ho sakta. Yeh binding-energy curve se match karta hai: calcium-40 (8.55 MeV per nucleon) neon-20 (8.03 MeV per nucleon) se zyada kaske bandha hai, aur sirf peak se kaafi aage ke nuclei hi tootne se fayda paate hain.
NumericalMediumNCERT Exercises 13.7
India ke planned thorium reactors uranium-233 jalayenge, jo per fission lagbhag 200 MeV release karta hai. (a) Agar 1.0 g uranium-233 ka har nucleus fission ho, to kitni energy release hogi? Jawaab MeV aur joules mein do. (b) Kitne tonnes coal ( $3.0 \times 10^{7} J$ per kg) utni hi energy dega?
J
Solution dikhao
$N = \frac{1.0 g}{233 g m o l^{- 1}} \times 6.022 \times 10^{23} m o l^{- 1} = 2.58 \times 10^{21}$
$E = 2.58 \times 10^{21} \times 200 M e V = 5.17 \times 10^{23} M e V = 8.3 \times 10^{10} J$
$m_{coal} = \frac{8.3 \times 10^{10} J}{3.0 \times 10^{7} J k g^{- 1}} = 2.8 \times 10^{3} k g$
Ek gram uranium, lagbhag 5 mm ka ek daana, lagbhag 3 tonnes coal ke barabar hai.
NumericalMediumNCERT Exercises 13.8
Ek fusion fuel pellet mein deuterium–tritium mixture ka 1.0 g hai, jismein D aur T atoms barabar ginti mein hain (atomic masses 2.014102 u aur 3.016049 u). Har D–T fusion 17.6 MeV release karta hai. Agar yeh poora fuse ho jaaye, to yeh energy 2.0 kW ka water heater kitni der chala sakti hai?
saal
Solution dikhao
Har fusion ek D aur ek T use karta hai, yaani $2.014102 + 3.016049 = 5.030151 u$ mass ka pair. 1.0 g mein pairs ki ginti:
$n = \frac{1.0 g}{5.030 g m o l^{- 1}} \times 6.022 \times 10^{23} = 1.197 \times 10^{23}$
$E = 1.197 \times 10^{23} \times 17.6 \times 1.602 \times 10^{- 13} J = 3.38 \times 10^{11} J$
$t = \frac{3.38 \times 10^{11} J}{2000 W} = 1.69 \times 10^{8} s \approx 5.3 saal$
NumericalMediumNCERT Exercises 13.9
Do alpha particles ko 1.9 fm radius ke hard spheres maano. Head-on chhoone ke liye unhe jo Coulomb barrier paar karna padega uski height MeV mein nikaalo. $1 / 4 π ε_{0} = 9 \times 10^{9} N m^{2} C^{- 2}$ aur $e = 1.6 \times 10^{- 19} C$ use karo.
MeV
Solution dikhao
Har alpha particle par charge $2 e$ hai; chhoote waqt unke centres $2 \times 1.9 = 3.8 f m$ door hain. Barrier us doori par electric potential energy hai:
$U = \frac{1}{4 π ε_{0}} \frac{(2 e) (2 e)}{r} = \frac{(9 \times 10^{9}) (4) (1.6 \times 10^{- 19})^{2}}{3.8 \times 10^{- 15}} J = 2.43 \times 10^{- 13} J$
$U = \frac{2.43 \times 10^{- 13}}{1.6 \times 10^{- 13}} M e V = 1.5 M e V$
Yeh do hydrogen nuclei ke comparison mein kai guna zyada hai, isiliye helium sirf boodhe hote stars ke kahin zyada garam cores mein hi jalta hai.
10
ProofMediumNCERT Exercises 13.10
$R = R_{0} A^{1 / 3}$ se shuru karke:
1. Dikhao ki nuclear matter ki mass density $A$ par depend nahi karti.
  Solution dikhao
  Nucleus ka mass lagbhag $A m_{u}$ hai aur uska volume $\frac{4}{3} π R_{0}^{3} A$ , isliye
  $ρ = \frac{A m_{u}}{\frac{4}{3} π R_{0}^{3} A} = \frac{3 m_{u}}{4 π R_{0}^{3}}$
  $A$ cancel ho jaata hai: har nucleus ki density same hai, lagbhag $2.3 \times 10^{17} k g m^{- 3}$ .
2. Nuclear matter ke har cubic femtometre mein kitne nucleons hain?
  per fm³
  Solution dikhao
  $n = \frac{A}{\frac{4}{3} π R_{0}^{3} A} = \frac{3}{4 π (1.2 f m)^{3}} = 0.138 f m^{- 3}$
3. Paas-paas ke nucleons ke beech typical doori ka estimate lagao aur use nuclear force ki range se compare karo.
  Solution dikhao
  Har nucleon ke hisse mein lagbhag $1 / n = 7.2 f m^{3}$ aata hai, yaani $n^{- 1 / 3} = 1.9 f m$ side ka cube. Padosi lagbhag 2 fm door hote hain, nuclear attraction ki range ke thoda andar aur uske repulsive core (0.8 fm) se bahar, jaisa saturation ke liye zaroori hai.
Quick answerEasy
Inmein se kaun se pairs isobars hain?
$_{}^{}_{19}^{39} K$ aur $_{}^{}_{19}^{41} K$ $_{}^{}_{6}^{13} C$ aur $_{}^{}_{7}^{14} N$ $_{}^{}_{6}^{14} C$ aur $_{}^{}_{7}^{14} N$ $_{}^{}_{1}^{2} H$ aur $_{}^{}_{2}^{4} H e$
Answer dekho
Answer (C) $_{}^{}_{6}^{14} C$ aur $_{}^{}_{7}^{14} N$
Isobars ka mass number same hota hai: carbon-14 aur nitrogen-14 dono ka $A = 14$ hai. Potassium wala pair isotopes hai, aur carbon-13 aur nitrogen-14 isotones hain ( $N = 7$ ).
Quick answerEasy
0.0100 u mass ke barabar kitni energy hai, MeV mein?
MeV
Answer dekho
Answer 9.315 MeV
$0.0100 \times 931.5 M e V = 9.32 M e V$ .
13
Quick answerMedium
In lengths ko sabse chhoti se sabse badi tak order mein lagao.
1. 7.7 MeV alpha ka gold se closest approach
2. gold atom ka radius
3. gold nucleus ka radius
4. proton ka radius (lagbhag 1 fm)
Answer dekho
Answer
proton ka radius (lagbhag 1 fm)
gold nucleus ka radius
7.7 MeV alpha ka gold se closest approach
gold atom ka radius
Lagbhag 0.9 fm, 7.0 fm, 30 fm aur $1.4 \times 10^{5} f m$ ( $0.14 n m$ ).
14
Quick answerEasy
Har decay ko us cheez se match karo jo woh nucleus ke saath karta hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$α$ decay : $Z$ 2 ghat-ta hai, $A$ 4
$β^{-}$ decay : $Z$ 1 badhta hai, $A$ nahi badalta
$β^{+}$ decay : $Z$ 1 ghat-ta hai, $A$ nahi badalta
$γ$ decay : $Z$ aur $A$ nahi badalte
Alpha particle 2 protons aur 2 neutrons le jaata hai; beta decay ek nucleon ko doosri tarah ke nucleon mein badalta hai; gamma decay sirf energy kam karta hai.
Quick answerEasy
Ek atomic mass unit MeV energy ke barabar hai (chaar significant figures tak).
Answer dekho
Answer 931.5
$(1 u) c^{2} = 1.4924 \times 10^{- 10} J = 931.5 M e V$ .
Quick answerEasy
True ya false: energy release karne wali fusion reaction mein products ka total mass reactants se kam hota hai.
SahiGalat
Answer dekho
Answer Sahi
Release hui energy $Q$ mass se aati hai: $\sum m_{initial} - \sum m_{final} = Q / c^{2} > 0$ .

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercises 13.1	Practice Q1
NCERT Exercises 13.2	Practice Q2
NCERT Exercises 13.3	Practice Q3
NCERT Exercises 13.4	Practice Q4
NCERT Exercises 13.5	Practice Q5
NCERT Exercises 13.6	Practice Q6
NCERT Exercises 13.7	Practice Q7
NCERT Exercises 13.8	Practice Q8
NCERT Exercises 13.9	Practice Q9
NCERT Exercises 13.10	Practice Q10

Quick check: MCQs

13.1 Introduction2 questions
Neutron ki discovery4 questions
13.3 Nucleus ka size3 questions
13.4.1 Mass – Energy2 questions
13.4.2 Nuclear binding energy4 questions
13.5 Nuclear force3 questions
13.6 Radioactivity3 questions
13.7.1 Fission3 questions
13.7.2 Nuclear fusion – stars mein energy generation3 questions
13.7.3 Controlled thermonuclear fusion2 questions

Glossary · 15

Key terms (English + Hindi)

atomic mass unitपरमाणु द्रव्यमान मात्रकparamanu dravyamaan maatrak: Carbon-12 atom ke mass ka barahvaan hissa: $1 u = 1.6605 \times 10^{- 27} k g$ , jo 931.5 MeV ke barabar hai.Chapter mein pehli baar yahan aaya →

binding energyबंधन ऊर्जाbandhan oorja: Nucleus ko free nucleons mein alag karne ke liye zaroori energy, jo uske bante waqt release hui energy ke barabar hai: $E_{b} = Δ M c^{2}$ .Chapter mein pehli baar yahan aaya →
binding energy per nucleonप्रति न्यूक्लिऑन बंधन ऊर्जाprati nyookliyon bandhan oorja: $E_{b} / A$ , ek nucleon ki average binding; zyaadatar nuclei ke liye lagbhag 8 MeV aur iron ke paas sabse zyada (8.8 MeV).Chapter mein pehli baar yahan aaya →

isobarसमभारिकsambhaarik: Aise nuclides jinka mass number $A$ same hai lekin $Z$ alag.Chapter mein pehli baar yahan aaya →
isotoneसमन्यूट्रॉनिकsamanyootronik: Aise nuclides jinka neutron number $N$ same hai lekin $Z$ alag.Chapter mein pehli baar yahan aaya →
isotopeसमस्थानिकsamasthanik: Ek hi element ke nuclides jo sirf neutrons ki ginti mein alag hain: $Z$ barabar, $N$ aur $A$ alag.Chapter mein pehli baar yahan aaya →

mass defectद्रव्यमान क्षतिdravyamaan kshati: Nucleus apne alag-alag protons aur neutrons se kitna halka hai: $Δ M = [Z m_{p} + (A - Z) m_{n}] - M$ .Chapter mein pehli baar yahan aaya →
mass numberद्रव्यमान संख्याdravyamaan sankhya: Nucleus mein nucleons ki total ginti, $A = Z + N$ .Chapter mein pehli baar yahan aaya →

nuclear fissionनाभिकीय विखंडनnaabhikiya vikhandan: Bhaari nucleus ka, jaise neutron absorb karne ke baad uranium-235 ka, do medium nuclei aur kuch neutrons mein tootna, jismein lagbhag 200 MeV release hota hai.Chapter mein pehli baar yahan aaya →
nuclear forceनाभिकीय बलnaabhikiya bal: Nucleons ko baandhne wala strong, short-range, charge-independent force; kuch femtometres ke baad yeh khatam ho jaata hai aur lagbhag 0.8 fm se neeche repel karta hai.Chapter mein pehli baar yahan aaya →
nuclear fusionनाभिकीय संलयनnaabhikiya sanlayan: Light nuclei ka judkar ek bhaari, zyada kaske bandha nucleus banana, jismein energy release hoti hai; yahi stars ko power deta hai.Chapter mein pehli baar yahan aaya →
nucleonन्यूक्लिऑनnyookliyon: Ek proton ya ek neutron.Chapter mein pehli baar yahan aaya →
nucleusनाभिकnaabhik: Atom ka chhota, dense, positively charged centre, jo protons aur neutrons se bana hai; ismein atom ka 99.9 % se zyada mass hota hai.Chapter mein pehli baar yahan aaya →

Q-valueQ-मानQ-maan: Nuclear process mein release hui energy: $(\sum m_{initial} - \sum m_{final}) c^{2}$ , jo kinetic energy mein hue gain ke barabar hai.Chapter mein pehli baar yahan aaya →

radioactivityरेडियोऐक्टिवताrediyoaiktivata: Unstable nuclei ka apne aap decay hona, jismein alpha particles, electrons ya positrons, ya gamma rays nikalti hain.Chapter mein pehli baar yahan aaya →

13.1 Introduction

13.2 Atomic masses aur nucleus ki composition

Neutron ki discovery

13.3 Nucleus ka size

13.4 Mass-energy aur nuclear binding energy

13.4.1 Mass – Energy

13.4.2 Nuclear binding energy

13.5 Nuclear force

13.6 Radioactivity

13.7 Nuclear energy

13.7.1 Fission

13.7.2 Nuclear fusion – stars mein energy generation

13.7.3 Controlled thermonuclear fusion

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye