Atoms: Class 12 Physics Chapter 12 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

85 min
1 simulation
14 practice questions
26 MCQs

Is chapter mein aap kya seekhenge

Samjho ki Geiger–Marsden ke α-scattering results ne Thomson ke model ko kaise reject kiya aur Rutherford ke chhote, dense, positive nucleus tak kaise pahunchaya
Impact parameter ko scattering angle se jodo aur energy conservation se distance of closest approach nikaalo
Dikhao ki circular orbit mein electron ke liye K = −E aur U = 2E hota hai, aur samjho ki negative total energy ka matlab bound hona kyun hai
Continuous, emission-line aur absorption spectra mein farak karo aur samjho ki line spectrum kisi element ko kaise pehchanta hai
Samjho ki classical physics Rutherford ke atom ke collapse ki prediction kyun karti hai, aur Bohr ke teen postulates batao
Hydrogen ke liye rₙ = n²a₀ aur Eₙ = −13.6/n² eV derive karo aur use karo, aur uska energy-level diagram padho (ground state, excitation, ionisation)
Kisi bhi transition mein photon ki frequency aur wavelength calculate karo aur hydrogen spectrum ki lines pehchano
Bohr ki quantisation condition ko standing de Broglie wave ke roop mein samjho aur Bohr model ki limitations batao

NCERT Physics Class 12, Chapter 12, pp. 290–305 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

12.1 Introduction

Raat mein low-pressure sodium lamps se roshan sadak par drive karo, to har car ek jaisi feeki orange-brown dikhti hai. Lamp ki light colours ka mix nahi hai: lagbhag saari light ek sharp wavelength par nikalti hai jo sirf sodium atoms dete hain, chahe lamp Pune mein ho ya kisi door ke star mein. Aisi wavelengths atom ke andar se aaye messages hain. Yeh chapter dikhata hai ki physicists ne inhe kaise padha: pehle thin metal foil par α-particles fire karke yeh pata lagaya ki atom apna charge aur mass kahan rakhta hai, phir hydrogen ke liye Niels Bohr ke bold rules se, jo uski light ko line by line predict karte hain.

Puzzle 1897 mein J. J. Thomson ki electron ki discovery se shuru hui. Unhone jitne bhi elements test kiye, sabne same chhota negative particle diya, jiska mass hydrogen atom ka lagbhag $1 / 1836$ hai. Kyunki poore atom par koi net charge nahi hota, usmein utna hi positive charge bhi hona chahiye. Aur kyunki electrons itne halke hain, atom ka lagbhag saara mass usi positive part ka hona chahiye. Positive part ki shape kya hai, aur atom ke andar electrons kahan baithe hain?

Thomson ka apna guess, 1898 mein diya gaya, sabse simple arrangement tha. Positive charge ko poore atom ke size ki ek ball mein barabar phaila do, aur electrons ko usmein aise points par laga do jahan un par forces balance ho jayein, jaise gol laddoo mein dabaye gaye kishmish. English writers ne ise plum pudding model kaha. Is picture mein mass bhi phaila hua hai, isliye atom mein kahin bhi koi hard core nahi hai (Fig. 12.1).

3/3Thomson ke atom se har α-particle lagbhag seedha nikal jaata hai. Nuclear atom mein woh rare particle jo nucleus ke paas se guzarta hai, side mein phenk diya jaata hai, ya wapas bhi.

Fig. 12.1Atom ki do pictures. Sirf ek chhota, dense, positive centre hi α-particle ko ghuma kar wapas bhej sakta hai. (Scale ke hisaab se nahi: asli nucleus drawing se hazaaron guna chhota hai.)NCERT p. 290

Same size ke do atoms saath-saath. Step 1: left atom, jis par Thomson model likha hai, plus signs se bhari ek green ball hai, jismein aath blue electrons jagah-jagah lage hain; right atom halka hai. Step 2: right atom, jis par Nuclear model likha hai, poora dikhta hai: ek dashed outer boundary, halki circular orbits jin par aath electrons hain, aur centre mein positive charge ka ek chhota sa dot. Step 3: orange alpha-particle tracks left se dono atoms ko cross karte hain. Thomson atom mein paanchon tracks lagbhag seedhe hain. Nuclear atom mein zyaadatar tracks seedhe hain, centre ke paas aim kiya gaya ek track tezi se neeche mudta hai, aur nucleus par lagbhag seedha aim kiya gaya ek track ghoom kar wapas left ki taraf jaata hai.

Light ne doosra clue diya. Hydrogen discharge tube ki chamak ko prism se guzaro, to sirf chaar visible colours dikhte hain, aisi wavelengths par jo ek saaf, regular pattern banati hain. 1885 mein ek Swiss schoolteacher, Johann Balmer, ne ek formula dhoondha jo inhe reproduce karta tha, bina yeh jaane ki woh kaam kyun karta hai. Har element ka apna lines ka pattern hota hai (Section 12.3). Iski tulna bulb ke filament se karo, jiski light mein har wavelength hoti hai. Farak spacing ka hai: thin gas mein har atom bina disturb hue radiate karta hai, isliye tumhe single atoms ki light dikhti hai, jabki solid mein bheed mein bhare atoms lagataar takraate aur energy share karte rehte hain, jo light ko ek continuous band mein phaila deta hai. Atom andar se jaisa bhi ho, use in single-atom patterns ko explain karna hi hoga.

Decisive evidence ek alag tarah ke experiment se aaya. Atoms ki nikali light study karne ke bajaye, Manchester mein Ernest Rutherford ke group ne radioactive source se tez α-particles ko thin metal foils se fire kiya aur record kiya ki woh kahan gaye (Section 12.2). Deflections ka pattern, jo Hans Geiger aur Ernest Marsden ne 1909 aur 1913 ke beech measure kiya, Thomson ke atom ke liye impossible tha. Iske liye zaroori tha ki atom ka lagbhag saara mass aur saara positive charge ek chhote central nucleus (नाभिक) mein ho, aur electrons bahut door bahar, Sun ke around planets ki tarah ghoomte hon. Yeh nuclear, ya planetary, model hai.

Planetary model ek bada kadam tha, lekin isne do mushkil sawaal chhod diye. Hydrogen, jo sabse simple atom hai, sharp wavelengths ki poori family kyun deta hai aur beech mein kuch nahi? Aur orbit mein ghoomta electron energy kho kar nucleus mein kyun nahi gir jaata? Classical physics kisi ka jawaab nahi deti (Section 12.4). Chapter ke doosre half mein Bohr ka model hydrogen ke liye dono ka jawaab deta hai.

Warm-up

2 mein se 1

Electron ka mass hydrogen atom ka lagbhag

1 / 1836

hai. Isse turant kya pata chalta hai?

Har atom mein lagbhag 1836 electrons hone chahiyeAtom ka zyaadatar charge electrons ke paas hota haiPositive charge ko poora atom bharna chahiyeAtom ka lagbhag saara mass uske positive part ka hota hai

Answer dekho

Answer (D) Atom ka lagbhag saara mass uske positive part ka hota hai

Electron ke mass ki tulna poore atom ke mass se karo.

12.2 Alpha-particle Scattering aur Rutherford ka Nuclear Model of Atom

α-particle (ऐल्फ़ा कण) helium atom ka nucleus hai: do protons aur do neutrons, charge $+ 2 e$ aur mass hydrogen atom ka lagbhag chaar guna. Radioactive sources inhe light ki speed ke lagbhag beesve hisse se phenkte hain, kuch MeV ki kinetic energy ke saath. α-particle ke liye gold leaf ki sheet atoms ka ek jungle hai, aur particles jungle se kaise bahar aate hain, isse pata chalta hai ki woh kis cheez ka bana hai.

Goal batana simple tha: gino ki kitne α-particles thin foil se apni original direction se har angle $θ$ par nikalte hain (Fig. 12.2). Counter zinc sulphide se coated ek chhoti screen thi, jahan bhi α-particle takraata wahan light ki ek halki flash, yaani scintillation, deti thi; ek observer use microscope se dekhta tha aur use foil ke around kisi bhi angle par ghuma sakta tha. Particles lead block mein dabaye ek radioactive source se aate the, jiska patla channel unki sirf ek patli pencil jaisi beam bahar aane deta tha. Target gold leaf tha jo ek metre ke millionth hisse se bhi patla tha, sirf kuch sau atoms gehra.Observers pehle lagbhag aadha ghanta andhere kamre mein baithte the taaki unki aankhein adjust ho jayein, phir ek baar mein sirf lagbhag ek minute ginte the, kyunki thakaan se accuracy jaldi gir jaati thi. Sab kuch vacuum mein rakha gaya taaki air molecules beam ko scatter na karein.

3/3 $150^{\circ}$ par bhi, foil ke source wali side par, kabhi-kabhi flash dikhti hai: lagbhag aath hazaar mein se ek α-particle $90^{\circ}$ se zyada mud jaata hai.

Fig. 12.2Alag-alag angles par α-particles ginna. Detector foil ke around ghoomta hai; angle badhne par count tezi se girta hai, lekin kabhi poora zero nahi hota.NCERT Fig. 12.2 · p. 292

Upar se dekhne par: left mein ek radioactive source patle slot wale lead block ke andar hai, aur ek orange beam right ki taraf ek thin gold foil tak jaati hai jo ek dashed circular track ke centre mein hai, jis par har 30 degrees par mark hai. Step 1: lagbhag seedhi orange rays ka ek fan foil ke aage ek detector tak jaata hai, jo zinc sulphide screen hai aur uske peeche microscope hai, seedha saamne 0 degrees par rakha; label par most here likha hai. Step 2: pehla detector fade hota hai aur detector 60 degrees par hai, ek deflected ray pakadta hai, label a few. Step 3: detector 150 degrees par hai, source wali side par, ek wapas aayi ray pakadta hai, label very rare. Ek note batata hai ki sab kuch vacuum ke andar hai.

Maan lo Thomson ki picture sahi hoti. To har gold atom charge ki ek soft, phaili hui ball hota jiska electric field kahin bhi bahut strong nahi hota. α-particle electron se lagbhag 7300 guna bhaari bhi hai, isliye electrons bhi use side mein nahi dhakel sakte. Har atom α-particle ko zyada se zyada degree ke chhote se hisse jitna hi hila sakta, aur sau atoms cross karne ke baad bhi total deflection ek-do degree ke andar hi rehna chahiye.

Counts ne kuch aur hi kaha. Zyaadatar α-particles sach mein lagbhag seedhe nikal gaye, lekin lagbhag saat sau mein se ek ek degree se zyada mud gaya, aur lagbhag aath hazaar mein se ek foil ki usi side se bahar aaya jahan se ghusa tha, yaani $90^{\circ}$ se zyada deflect hua (Fig. 12.3). Rutherford ne baad mein kaha ki yeh utna hi chaunkane wala tha jitna tissue paper ki sheet se bhaari artillery shell ka wapas uchhal jaana.

3/3Geiger aur Marsden ke counts ten ki paanch se zyada powers tak nuclear curve ke saath chale, rare bade angles ke saath.

Fig. 12.3Har angle par counts (

180^{\circ}

wale count ke relative). Nucleus kuch α-particles ko har angle par scatter karta hai; Thomson ka atom kuch degrees ke baad kisi ko scatter nahi karta.NCERT Fig. 12.3 · p. 293

Counts ka graph, logarithmic scale par 1 se 100 000 tak, scattering angle ke against 0 se 180 degrees tak. Step 1: nuclear atom label wala solid blue curve top left se tezi se girta hai, 20 degrees par lagbhag 1000 aur 90 degrees par 4 se hokar, phir 180 degrees par 1 tak flat ho jaata hai. Step 2: Thomson atom label wala dashed vermillion curve top se lagbhag vertical girta hai aur 5 degrees se pehle hi graph ke bottom se bahar gayab ho jaata hai. Step 3: 90 degrees ke aage wala region shaded hai aur us par rare, not zero likha hai.

Andaaza lagao ki har tarah ka atom kitna zor se dhakel sakta hai. Gold atom ke $+ 79 e$ ko $10^{- 10} m$ radius ki ball mein phaila do, to sabse strong field, uski surface par, lagbhag $10^{13} N / C$ hota hai. Poori ball cross karne par α-particle ki energy sirf lagbhag ek kiloelectronvolt badalti, jabki uske paas kai MeV hain, isliye use mushkil se pata chalta. Usi charge ko $7 f m$ radius ki ball mein daba do, to kyunki field $1 / r^{2}$ ki tarah chalta hai, uski surface par field lagbhag $(10^{- 10} / 7 \times 10^{- 15})^{2} \approx 2 \times 10^{8}$ guna zyada strong hota hai. Sirf aisa concentrated, bhaari centre hi α-particle ko ghuma kar wapas bhej sakta hai. Rutherford ne calculate kiya ki point-like nucleus har angle par kitne α-particles bhejega. Predicted count angle ke saath smoothly lekin tezi se giraPoint nucleus ke liye screen ke per unit area count $1 / \sin^{4} (θ / 2)$ ke proportional hota hai. Yeh formula syllabus se bahar hai., aur Geiger aur Marsden ke 1913 ke counts $5^{\circ}$ se $150^{\circ}$ tak isi ke saath chale.

To α-particle ki nazar se gold foil chhote, hard dhabbon ka ek patla kohra hai. Baad ke measurements ne nuclear radii lagbhag $1 f m$ aur $7 f m$ ke beech paaye ( $1 f m = 10^{- 15} m$ , ek femtometre ya fermi), jabki poora atom lagbhag $10^{- 10} m$ ka hai, yaani yeh dhabbe apni jagah se das hazaar se ek lakh guna chhote hain (Example 12.1). Kohra cross karta particle aam taur par har dhabbe ko miss karta hai aur lagbhag bina disturb hue nikal aata hai. Jo rare particle kisi dhabbe ke bahut paas se guzarta hai, woh side mein, ya wapas, phenk diya jaata hai.

Definition

Nucleus

Atom ka chhota, dense, positively charged centre, jismein uska saara positive charge aur 99.9 % se zyada mass hota hai. Iska radius kuch femtometres hota hai, atom ke radius ka lagbhag $10^{- 4}$ se $10^{- 5}$ .

ExampleGold nucleus ka charge $+ 79 e$ aur radius lagbhag $7 f m$ hota hai.

Path predict karne ke liye teen idealisations chahiye, aur har ek experiment se support hota hai:

Nucleus apni jagah rehta hai. Gold nucleus (mass $197 u$ ) α-particle ( $4 u$ ) se lagbhag pachaas guna bhaari hai, isliye uske recoil ko neglect kar sakte hain.
Electrons ka koi role nahi. Woh α-particle se hazaaron guna halke hain.
Zyada se zyada ek close encounter. Foil itni patli hai ki α-particle shayad hi kabhi ek se zyada nucleus ke paas se guzarta hai.

Inke saath, α-particle par sirf ek force hai, charge $Z e$ wale nucleus ka Coulomb repulsion (gold ke liye $Z = 79$ ):

Nucleus ke paas α-particle par force

F = \frac{1}{4 π ε_{0}} \frac{(2 e) (Z e)}{r^{2}}

(12.1)

jahan

$r$: α-particle aur nucleus ke centres ke beech ki distance
$Z e$: nucleus ka charge ( $Z$ atomic number hai)
$2 e$: α-particle ka charge

Force dono ko jodne wali line ke along lagta hai, aur jaise-jaise α-particle paas aata hai aur phir door jaata hai, uska size aur direction dono lagataar badalte hain. Is force ke saath Newton ka second law path deta hai: hyperbola ki ek branch, jo nucleus se door mudti hai.

Try karke dekho

Ek target dhoondho jo tum dekh nahi sakte

Tumhe chahiye

lagbhag 50 cm chaudi ek stiff board ya tray
barabar motai ki aath books
lagbhag 8 cm across ek gol tin
20 marbles
ek ruler

Pehle guess karo

Tin board ki chaudai ka lagbhag paanchva hissa block karta hai. Barabar doori wali lanes mein roll ki gayi 20 marbles mein se, tumhare hisaab se lagbhag kitni side mein dhakeli jayengi?

Ek bhi nahiLagbhag 4Lagbhag 10Saari 20

Lagbhag 50 cm chaudi ek stiff board ko books ke chaar barabar stacks par rakho taaki marble uske neeche roll kar sake.
Jab tum doosri taraf dekh rahe ho, ek partner lagbhag 8 cm across ek gol tin board ke neeche kahin chhupa de.
Ek edge ke along 5 cm ki doori par das launch lanes mark karo. Har lane ke along do marbles seedhi, same dheemi speed se roll karo.
Har marble ke liye note karo ki woh seedhi nikli ya side mein dhakeli gayi, aur lagbhag kis direction mein.
Deflected marbles gino. Deflected fraction ko board ki chaudai se multiply karne par tin ki chaudai plus ek marble ki chaudai ka andaaza milta hai.

Table 12.1Kya har marble deflect hui? (yes ya no)

	Marble 1	Marble 2
Lane 1 (0 cm)
Lane 2 (5 cm)
Lane 3 (10 cm)
Lane 4 (15 cm)
Lane 5 (20 cm)
Lane 6 (25 cm)
Lane 7 (30 cm)
Lane 8 (35 cm)
Lane 9 (40 cm)
Lane 10 (45 cm)

Tumhari entries apne aap save hoti hain.

Aam taur par ek ya do lanes tin ko cross karti hain, isliye 2 ya 4 marbles side mein dhakeli jaati hain aur baaki seedhi roll ho jaati hain. Har lane board ki 5 cm strip ko represent karti hai, isliye tin ki chaudai lagbhag (hit hui lanes × 5 cm) minus ek marble ki chaudai hai; estimate ko behtar karne ke liye lanes ko 2.5 cm shift karke dohrao. Tumne tin ko bina dekhe uska size naap liya. Rutherford ne α-particles ke saath yahi kiya: bade deflections ke rare hone se unhe pata chala ki nucleus kitna chhota hai. Ek farak important hai: marble sirf chhoone par uchhalti hai, jabki α-particle khaali space ke across lagne wale force se mudta hai, isliye uska path smoothly mudta hai.

12.2.1 Alpha-particle trajectory

Sabse simple case se shuru karo: nucleus ke bilkul centre par aim kiya gaya α-particle. Woh electric 'hill' chadhte hue dheema hota hai, ek pal ke liye rukta hai aur apni hi line par seedha wapas dhakela jaata hai, isliye $θ = 180^{\circ}$ . Ab aim ko side mein shift karo. Particle ki original line aur nucleus ke centre ke beech ki perpendicular distance ko impact parameter $b$ (संघट्ट प्राचल) kehte hain, aur incoming aur outgoing directions ke beech ka angle scattering angle $θ$ hai.

Definition

Impact parameter

Particle ki initial line of motion aur target nucleus ke centre ke beech ki perpendicular distance $b$ .

ExampleHead-on collision mein $b = 0$ hota hai.

Jab $b$ zero se thoda hi zyada ho, particle nucleus ke around ghoom kar bade angle par nikalta hai. Bade $b$ par woh door rehta hai, kamzor push mehsoos karta hai aur kam mudta hai; kaafi door ho to use nucleus ka pata hi nahi chalta (Fig. 12.4).

3/4 $b = d / 4$ : particle nucleus ke around ghoomta hai, $θ \approx 127^{\circ}$ .

Fig. 12.4Chaar impact parameters ke liye Coulomb's law se compute kiye gaye paths. Distances

d

ki units mein hain, jo is energy ka α-particle head-on collision mein sabse paas aa sakta hai (Example 12.2).NCERT Fig. 12.4 · p. 294

Ek nucleus dashed horizontal axis par hai, aur alpha-particles left se axis ke upar aate hain. Step 1: impact parameter b = 2d wala particle dheere se door mudta hai aur lagbhag 28 degrees par nikalta hai; left mein ek bracket b ko mark karta hai. Step 2: b = d wala particle lagbhag 53 degrees par nikalta hai; pehla path fade hota hai. Step 3: b = d ka ek chauthai wala particle nucleus ke around ghoomta hai aur lagbhag 127 degrees par nikalta hai, upar aur wapas left ki taraf jaate hue. Step 4: nucleus par seedha aim kiya gaya particle distance d par rukta hai aur apni hi line par 180 degrees par wapas aata hai.

Lekin beam aim nahi ki jaati. Uske saare particles ki energy lagbhag same hoti hai lekin woh foil par random positions par aate hain, aur lagbhag sabke impact parameters figure wale kisi bhi impact parameter se kahin bade hote hain. Sirf woh particles jo kisi nucleus ke around ek chhoti si disc se guzar jaate hain, bade angles par mudte hain.Fixed nucleus ke liye $\tan (θ / 2) = d / 2 b$ , isliye sirf $b < d / 2$ wale particles hi $90^{\circ}$ se zyada mudte hain. Tum neeche wale lab mein yeh check kar sakte ho. Isliye wapas uchhalne wala fraction us disc ka size naapta hai, aur isse pata chalta hai ki charged centre kitna chhota hai. Yeh nucleus ke size ki upper limit deta hai: agar sabse paas aane wale α-particles bhi Coulomb's law follow karte hain, to nucleus unki distance of closest approach se chhota hona chahiye.

InteractiveScattering lab

Panel ke centre mein ek nucleus hai, aur ek dashed circle dikhata hai ki alpha-particle use kahan touch karega. Tum foil chunte ho (gold, silver, copper ya aluminium), alpha-particle ki energy 1 se 10 MeV tak aur uska impact parameter 0 se 200 femtometres tak. Computed path draw hota hai, b dikhane wale bracket ke saath, aur readouts head-on closest distance d, scattering angle aur is path par closest approach dete hain. Ek box b ke liye d ke chauthai se d ke chaar guna tak paths ki family draw kar sakta hai. Fire button alpha-particle ko uske path par chalata hai, aur ek message batata hai ki woh wapas phenka gaya ya nahi aur nucleus ko touch karega ya nahi.

Fig. 12.5Ek foil, ek energy aur ek impact parameter chuno, phir α-particle ko nucleus ke paas se bhejne ke liye Fire dabao.

Example 12.1NCERT Example 12.1 · p. 294

Atom kitna khaali hai?

Hydrogen atom mein proton ka radius lagbhag $0.84 f m$ hai aur electron aam taur par usse $5.3 \times 10^{- 11} m$ door hota hai. (a) Socho ki poore atom ko itna bada kar diya jaaye ki proton cricket ball (radius $3.6 c m$ ) jitna ho jaaye. Electron kitni door hoga? (b) Earth ka radius $6.37 \times 10^{6} m$ hai aur Moon uske around $3.84 \times 10^{8} m$ par ghoomta hai. Kaun zyada khaali hai, hydrogen atom ya Earth–Moon system, aur lagbhag kitne factor se?

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

Volume ke hisaab se contrast aur bhi tez hai: proton electron ki orbit ke andar wale sphere ka sirf lagbhag $4 \times 10^{- 15}$ bharta hai. Solar system bhi, jo hum mein se zyaadatar ke liye sabse khaali arrangement hai, iske muqable bhara hua hai, jaisa neeche wala simulation dikhata hai.

SimulationSolar system kitna khaali hai?

Sun aur chaar inner planets, Mercury, Venus, Earth aur Mars, apni orbits par, sizes aur distances ke true ratios ke saath drawn, isliye Sun aur planets chhote dots ban jaate hain. Play dabane par planets Sun ke around chalne lagte hain. Ek switch schematic view mein badalta hai jismein sizes exaggerate kiye gaye hain taaki planets dikh sakein.

Fig. 12.6Inner planets true scale par drawn hain. Unhe chalane ke liye Play dabao, aur squeezed schematic view se tulna ke liye scale switch use karo.

Try karo

True scale par Sun ke dot ki tulna Earth ki orbit ki chaudai se karo: radii ka ratio lagbhag 1 aur 216 hai.
Schematic view par switch karo. Planets ko dikhane ke liye kya exaggerate karna pada?
Agar hydrogen atom ko itna bada kiya jaaye ki uska nucleus is Sun jitna ho, to uska electron Earth se lagbhag 290 guna door hoga: Neptune se lagbhag das guna door.

Example 12.2NCERT Example 12.2 · p. 295

Lead nucleus tak closest approach

Polonium-210 $5.30 M e V$ kinetic energy wale α-particles nikaalta hai. Unmein se ek thin lead foil mein ek lead nucleus ( $Z = 82$ ) ki taraf seedha jaata hai. (a) Rukne aur wapas mudne se pehle woh kitna paas aata hai? (b) Lead nucleus aur α-particle ke radii lagbhag $7.1 f m$ aur $1.7 f m$ hain. Nucleus ko bas touch karne ke liye α-particle ko kitni kinetic energy chahiye? Lead nucleus ko fixed maano: woh lagbhag 52 guna bhaari hai.

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

To Rutherford ke experiments mein α-particles ne kabhi gold ya lead nucleus ko touch nahi kiya. Woh sirf electric force ke kaaran usse kaafi bahar hi mud gaye. Isiliye counts ne Coulomb's law ko itni wafadaari se follow kiya, aur isiliye scattering nucleus ka size naap nahi saki, sirf uski upper limit set kar saki.

12.2.2 Electron orbits

Rutherford ke model ko abhi bhi batana hai ki electrons kya kar rahe hain. Woh ruke nahi reh sakte, warna nucleus unhe seedha andar kheench lega. Isliye planets ki tarah unhe orbits mein chalna hoga, jahan Coulomb attraction woh centripetal force deta hai jo unhe mudta rakhta hai. Sabse simple atom, hydrogen, lo: charge $- e$ aur mass $m$ wala ek electron, jo proton ke around radius $r$ ke circle mein speed $v$ se chal raha hai.

DerivationCircular orbit mein electron ki energy

Circular orbit ke liye Coulomb attraction mass times centripetal acceleration ke barabar hona chahiye:

\frac{1}{4 π ε_{0}} \frac{e^{2}}{r^{2}} = \frac{m v^{2}}{r}

(12.2)

Rearrange karne par yeh orbit ke radius ko usmein speed se jodta hai. Tez electron zyada paas ghoomta hai:

r = \frac{e^{2}}{4 π ε_{0} m v^{2}}

(12.3)

(12.2) ko $r / 2$ se multiply karne par seedhe kinetic energy milti hai. Electron–proton pair ki potential energy door hone par zero hai aur yahan negative hai, kyunki woh attract karte hain:

K = \frac{1}{2} m v^{2} = \frac{e^{2}}{8 π ε_{0} r}, U = - \frac{e^{2}}{4 π ε_{0} r}

Dono ko jodne par total energy milti hai:

E = K + U = - \frac{e^{2}}{8 π ε_{0} r}

(12.4)

Do results yaad rakhne layak hain (Fig. 12.7). Pehla, $K = - E$ aur $U = 2 E$ : potential energy size mein kinetic energy ki do guna hai, aur sign mein ulti. Doosra, total energy negative hai. Electron ko infinity tak kheench kar wahan rest mein chhodne ke liye tumhe kam se kam $∣ E ∣$ dena hoga, isliye tab tak woh proton se bound hai. Positive total energy ke saath electron extra energy ke saath infinity tak pahunch sakta: woh proton ke paas se ek baar ghoom kar chala jaata, us comet ki tarah jo kabhi wapas nahi aata.

3/3Total $E = K + U = - K$ negative hai: electron bound hai.

Fig. 12.7Circular orbit mein energies aapas mein locked hain:

U = - 2 K

, isliye

E = K + U = - K

.NCERT p. 296

Left mein ek dashed circular orbit proton ko ghere hue hai, orbit par ek electron hai, uska velocity arrow orbit ke along hai aur radius r uss tak drawn hai. Right mein ek horizontal zero line. Step 1: height K wala green bar line ke upar khada hai. Step 2: usse do guna lamba vermillion bar line ke neeche latka hai, label U equals minus 2K. Step 3: green jitna lamba blue bar line ke neeche latka hai, label E equals minus K, aur ek dashed guide dikhati hai ki woh U se theek aadha neeche tak jaata hai.

Example 12.3NCERT Example 12.3 · p. 296

Binding energy se size aur speed

Ek khaas excited state mein hydrogen atom se electron nikaalne mein $1.51 e V$ lagta hai. Orbit ko circle maan kar nikaalo (a) uska radius, (b) electron ki speed aur (c) uski kinetic aur potential energies.

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

Quick check

3 mein se 1

Proton ke around circular orbit mein ek electron ki kinetic energy

3.40 e V

hai. Uski total energy kya hai?

Answer dekho

Answer -3.4 eV

Circular Coulomb orbit mein

E = - K

, isliye

E = - 3.40 e V

. (Uski potential energy

U = 2 E = - 6.80 e V

hai.)

12.3 Atomic Spectra

Prism ya grating light ko wavelength ke hisaab se spectrum mein phaila deta hai, aur Fig. 12.8 ke teen tarah ke spectrum teen alag kahaniyan sunate hain. Bulb ka chamakta filament colour ka ek bina toota band deta hai, yaani continuous spectrum. Discharge tube mein low pressure par hydrogen, jo current guzarne se chamakti hai, beech mein andhere ke saath sirf kuch bright, narrow lines deti hai: emission line spectrum (उत्सर्जन रेखीय स्पेक्ट्रम).

Definition

Emission line spectrum

Low pressure par excited gas ya vapour ki light mein sharp bright lines ka set, jo element ki characteristic wavelengths par hoti hain.

ExampleSodium vapour lagbhag poori tarah $589 n m$ ke aas-paas chamakti hai, sodium street lamps ka peela rang.

Har element ka lines ka apna arrangement hota hai, packet par barcode jitna alag. Sample ki ek chutki ko flame ya plasma mein garam karo, uski lines padho, aur tumhe pata chal jaata hai ki usmein kaun se elements hain, itni kam maatra mein bhi jo taula na ja sake; laboratories paani aur mitti mein lead ya arsenic aise hi test karti hain. Astronomers inhi lines se pata lagate hain ki arabon kilometres door ke stars kis cheez ke bane hain.

Teesri kahani ulti chalti hai. White light ko thandi hydrogen se bhejo aur jo bahar aata hai use dekho. Lagbhag har colour bach jaata hai, lekin patle dark gaps dikhte hain, aur har gap theek wahin hai jahan hydrogen ki bright emission line hoti hai. Lagta hai gas guzarti light se theek wahi wavelengths hata deti hai jo woh chamakte waqt nikaalti. Gaps ka yeh pattern gas ka absorption spectrum (अवशोषण स्पेक्ट्रम) hai; Bohr ka model samjhayega ki dono sets match kyun karte hain (Section 12.5).

Definition

Absorption spectrum

Dark lines se kata continuous spectrum, jo tab banta hai jab white light ek thandi gas se guzarti hai jo apni characteristic wavelengths absorb kar leti hai.

ExampleSunlight mein dark lines, jo Sun ki thandi outer layers ki gases banati hain.

4/4Gayab wavelengths theek wahi hain jo emit hoti hain.

Fig. 12.8Continuous, emission aur absorption spectra. Hydrogen theek wahi wavelengths absorb karti hai jo woh emit karti hai.NCERT Fig. 12.5 · p. 297

Teen horizontal strips ka 400 se 700 nanometres tak ek hi wavelength scale hai. Step 1: top strip, label hot filament: every colour, ek smooth band hai jo violet se blue, green aur yellow hote hue red tak jaata hai. Step 2: middle strip, label glowing hydrogen: bright lines, plain hai, sirf chaar narrow coloured lines hain, violet lagbhag 410 ke paas, blue lagbhag 434 ke paas, sky blue lagbhag 486 ke paas aur red lagbhag 656 nanometres ke paas. Step 3: bottom strip, label through cool hydrogen: gaps, phir se smooth band hai jismein chaar patle gaps kate hain. Step 4: dashed vertical guides teeno strips se guzarti hain, aur dikhati hain ki gaps theek bright lines ke neeche hain.

Try karke dekho

Purani CD se spectroscope

Tumhe chahiye

ek purani CD ya DVD jo tum de sako
ek chhota cardboard box, jaise toothpaste ka carton
kainchi aur tape
lamps: filament bulb, tube light ya CFL, white LED, aur raat mein sodium street lamp

Pehle guess karo

Tumhare hisaab se kaunsa lamp smooth band ke bajaye alag-alag bright lines dikhayega?

Purana filament bulbCandle ki flameTube light ya CFLSunlight mein white wall

Kisi bade se purani CD ya DVD ka lagbhag 3 cm square tukda katwao. Kate edges ko dhyaan se pakdo.
Chhote box (toothpaste carton chalega) ke ek end mein lagbhag 1 mm chauda patla slit kaato, aur doosre end ke paas top mein dekhne ka ek hole.
CD ke tukde ko box ke andar viewing hole ke neeche tape karo, shiny side upar, itna tilt karke ki slit ki light us par pade.
Slit ko baari-baari har lamp ki taraf karo, hole se dekho, aur box ko tab tak tilt karo jab tak colours ka band na dikhe.
Record karo ki tumhe smooth band dikhta hai ya alag lines, aur kaun se colours.

Table 12.2Har lamp ki light mein kya hai?

	Smooth band ya lines?	Dikhne wale colours
Filament bulb
Tube light ya CFL
White LED
Peela sodium street lamp
Candle ki flame

Tumhari entries apne aap save hoti hain.

Filament bulb aur candle smooth band dete hain, kyunki garam solids (filament, flame mein chamakti kaalikh) har wavelength radiate karte hain. Sodium lamp apni lagbhag saari light ek bright yellow line mein deta hai. Tube light apne mercury vapour se alag violet, blue aur green lines dikhati hai, saath mein uski coating ki light bhi. White LED ek broad blue peak aur ek broad yellow band deta hai, lines nahi balki smooth, kyunki uski light solids se aati hai.

12.4 Hydrogen Atom ka Bohr Model

Electromagnetic Waves mein tumne dekha ki koi bhi accelerated charge radiate karta hai; radio antenna aise hi broadcast karta hai. Circular orbit mein electron hamesha accelerate kar raha hai, kyunki uski velocity ki direction badalti rehti hai, isliye isi rule se use apni frequency of revolution par lagataar broadcast karna chahiye, aur aisa karte hue energy khoni chahiye. Planets is anjaam se bach jaate hain kyunki un par koi net charge nahi hota; electrons nahi bach sakte.

Ab equations ko ek ke baad ek jodo (Fig. 12.9). Radiate karne se $E$ aur negative hota hai; (12.4) se zyada negative $E$ matlab chhota $r$ ; (12.3) se chhota $r$ matlab bada $v$ ; aur isliye revolution frequency $v / 2 π r$ badhti hai. To electron aur tezi se andar ki taraf ghoomta jaata hai, uska broadcast frequency mein lagataar upar khisakta hai, jab tak woh nucleus tak na pahunch jaaye. Real atoms is kahani ke dono hisson ko galat saabit karte hain: woh stable hain, aur smear ke bajaye sharp lines emit karte hain.

2/3Energy khote hue electron spiral karke andar jaata hai; uska revolution, aur uski nikali light, tez aur tez hote jaate hain.

Fig. 12.9Rutherford ke atom ke liye classical physics kya predict karti hai. Real atoms stable hain aur sharp lines dete hain, isliye classical picture mein kuch na kuch fail hona chahiye.NCERT Fig. 12.6 · p. 298

Left mein proton aur blue circular orbit par ek electron; right mein emitted wave wala ek horizontal axis. Step 1: electron fixed orbit par ghoomta hai aur right mein wave ek steady sine hai, label one steady frequency. Step 2: orbit ki jagah ek blue spiral aata hai jo andar proton tak ghoomta hai, aur right mein wave ek chirp ban jaati hai jiske crests paas aur paas aate jaate hain, label frequency keeps rising. Step 3: electron proton tak pahunch gaya, spiral fade hota hai, aur label atom collapses lagbhag 10 to the minus 11 seconds ke time ke saath dikhta hai.

Example 12.4NCERT Example 12.4 · p. 298

Orbit ki classical frequency

Example 12.3 wala electron lo, orbit radius $4.77 \times 10^{- 10} m$ aur speed $7.29 \times 10^{5} m / s$ . Classical electromagnetism ke hisaab se woh kis frequency par radiate karna shuru karega, aur woh spectrum ke kis part mein hai? Electron ke spiral karke andar jaane par is frequency ka kya hoga?

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

1913 mein Bohr ka jawaab practical tha. Jo kaam karta hai use rakho, orbit ke liye Newton's laws aur Coulomb's law, aur Planck aur Einstein ke quantum ideas se inspired teen rules jodo, jo atom ke andar classical electromagnetism ko seedha overrule kar dete hain. Woh rules ko justify nahi kar sake; woh sirf unhe test kar sakte the, aur hydrogen ka spectrum hi test tha.

PostulateBohr ka pehla postulate: stationary states

Kuch khaas orbits hain jinmein electron radiate nahi karta, classical electromagnetism chahe jo kahe. Jab tak uska electron unmein se kisi ek mein hai, atom ki total energy fixed rehti hai; aisi state ko stationary state (स्थायी अवस्था) kehte hain.

Bohr ne iska koi mechanism nahi diya. Unhone bas declare kiya ki radiation law in khaas orbits par laagu nahi hota, aur predictions ko faisla karne diya. Unka doosra rule batata hai ki kaun si orbits khaas hain.

PostulateBohr ka doosra postulate: quantised angular momentum

Allowed orbits woh hain jinmein electron ka angular momentum $h / 2 π$ ka whole-number multiple ho, jahan $h$ Planck's constant hai.

Bohr ki quantisation condition

L = m v r = \frac{n h}{2 π}, n = 1, 2, 3, \dots

(12.5)

jahan

$h$: Planck's constant, $6.63 \times 10^{- 34} J s$
$n$: ek positive whole number, principal quantum number

Saari quantities mein se angular momentum hi kyun? $h$ ki units check karo: $J s = k g m^{2} / s$ , theek angular momentum $m v r$ ki units. Circular orbit ke liye angular momentum quantise karne ki natural quantity hai, aur $h / 2 π = 1.055 \times 10^{- 34} J s$ natural step size hai. Whole number $n$ ko principal quantum number (मुख्य क्वांटम संख्या) kehte hain.

PostulateBohr ka teesra postulate: har jump mein ek photon

Light tabhi nikalti hai jab electron ek stationary state se kam energy wali state mein jump karta hai. Ek photon poora energy difference le jaata hai, isliye uski frequency $ν$ jump se pehle ki energy $E_{i}$ aur baad ki energy $E_{f}$ se tay hoti hai.

Emitted photon ki frequency

h ν = E_{i} - E_{f}, E_{i} > E_{f}

(12.6)

Ulta chalane par yahi rule absorption batata hai: atom photon ko tabhi le sakta hai jab $h ν$ kisi higher state tak ke gap se theek match kare. Dhyaan do ki light ki frequency kya nahi hai. Yeh kisi bhi orbit ki frequency of revolution nahi hai, jo classical expectation thi.

Postulates allowed orbits ko poori tarah fix kar dete hain. Orbit condition (12.2) ko quantisation rule (12.5) ke saath jodne par hydrogen ki har stationary state ka radius, speed aur energy milta hai.

Derivationnth orbit ka radius, speed aur energy

(12.5) se allowed orbit mein speed $v = n h / 2 π m r$ hai. Ise $m v^{2} = e^{2} / 4 π ε_{0} r$ mein daalo, jo (12.2) ko $r$ se multiply karke milta hai:

m {(\frac{n h}{2 π m r})}^{2} = \frac{e^{2}}{4 π ε_{0} r}

$r$ ke liye solve karne par allowed radii milte hain:

r_{n} = \frac{n^{2}}{m} {(\frac{h}{2 π})}^{2} \frac{4 π ε_{0}}{e^{2}} = \frac{ε_{0} h^{2}}{π m e^{2}} n^{2}

(12.7)

Quantisation rule ko $m r_{n}$ se divide karne par $n$ th orbit mein speed milti hai:

v_{n} = \frac{n h}{2 π m r_{n}} = \frac{e^{2}}{2 ε_{0} h} \frac{1}{n}

Aakhir mein $r_{n}$ ko (12.4) mein daalo:

E_{n} = - \frac{e^{2}}{8 π ε_{0} r_{n}} = - \frac{m e^{4}}{8 ε_{0}^{2} h^{2}} \frac{1}{n^{2}}

(12.8)

$m = 9.11 \times 10^{- 31} k g$ , $e = 1.602 \times 10^{- 19} C$ , $ε_{0} = 8.854 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}$ aur $h = 6.626 \times 10^{- 34} J s$ ke saath constants dete hain

E_{n} = - \frac{2.18 \times 10^{- 18} J}{n^{2}}

(12.9)

ya, $1.602 \times 10^{- 19} J / e V$ se divide karke,

E_{n} = - \frac{13.6 e V}{n^{2}}

(12.10)

Wahi constants radii aur speeds dete hain:

r_{n} = (5.29 \times 10^{- 11} m) n^{2}, v_{n} = \frac{2.19 \times 10^{6} m / s}{n}

Sabse chhoti orbit, $r_{1} = 5.29 \times 10^{- 11} m$ , Bohr radius $a_{0}$ (बोर त्रिज्या) hai. Yeh kinetic theory of gases se jaane gaye hydrogen atom ke size se match karta hai, model ki pehli success. Radii $n^{2}$ ki tarah badhte hain (Fig. 12.10) aur speeds $1 / n$ ki tarah girti hain. Sabse andar wali orbit mein bhi electron light ki speed ke 1 % se kam par chalta hai, isliye relativity ko ignore karna safe hai.

3/3 $r_{4} = 16 a_{0}$ , angular momentum $4 h / 2 π$ ke saath.

Fig. 12.10Hydrogen ki Bohr orbits scale par.

n

mein har ek step bahar jaane par angular momentum mein

h / 2 π

ki ek unit judti hai, lekin radius

n^{2}

ki tarah badhta hai.NCERT p. 299

Proton ke around concentric blue circles, scale par drawn, jahan har circle right ki taraf ek dashed ray ko cross karta hai wahan ek electron dot hai aur uske neeche n ki value likhi hai. Step 1: n equals 1 orbit proton ke paas ek chhota circle hai aur n equals 2 orbit chaar guna chaudi hai. Step 2: n equals 3 orbit dikhti hai, pehli ki nau guna. Step 3: n equals 4 orbit dikhti hai, pehli ki solah guna, jo drawing ka zyaadatar hissa bhar deti hai. Bagal ke formulas batate hain ki radius n squared times a zero ke barabar hai aur angular momentum n h over 2 pi ke barabar hai.

Example

n = 2 orbit numbers mein

Hydrogen ki $n = 2$ state ke liye orbit radius, electron ki speed, ek revolution ka time aur total energy nikaalo.

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

Quick check

4 mein se 1

Bohr ke har postulate ko uski baat se match karo.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

Pehla postulate : Kuch orbits radiate nahi karti
Doosra postulate : $L = n h / 2 π$
Teesra postulate : $h ν = E_{i} - E_{f}$

Pehla postulate stable, non-radiating orbits allow karta hai; doosra angular momentum ko quantise karke unhe chunta hai; teesra kehta hai ki light sirf unke beech jumps mein emit hoti hai.

12.4.1 Energy levels

Equation (12.10) woh har energy batati hai jo hydrogen atom ki ho sakti hai jab uska electron bound hai: $- 13.6$ , $- 3.40$ , $- 1.51$ , $- 0.85 e V$ waghera. Inhe vertical energy scale par horizontal lines ki tarah draw karne se energy-level diagram banta hai (Fig. 12.11). Ise neeche se upar padho.

Diagram ko ek seedhi samjho. Sabse neeche wala danda, $n = 1$ , ground state (निम्नतम अवस्था) hai, jismein sabse tightly bound electron radius $a_{0}$ wali orbit mein hai. Akela chhodne par har hydrogen atom wahin pahunch jaata hai, isiliye room temperature par hydrogen ka jar lagbhag poora ground state mein hota hai. Us dande se seedhi ke top ke bhi paar, $E = 0$ tak, chadhne mein $13.6 e V$ lagta hai: yeh ionisation energy (आयनन ऊर्जा) hai. Real hydrogen par experiments bhi yahi figure dete hain, jo model ki sabse convincing successes mein se ek hai.

Definition

Ground state

Atom ki sabse kam energy wali state. Hydrogen ke liye yeh $n = 1$ hai, $E_{1} = - 13.6 e V$ ke saath, aur electron radius $a_{0}$ wali orbit mein hota hai.

Definition

Ionisation energy

Ground state wale atom se electron ko nikaal kar free aur rest mein chhodne ke liye zaroori sabse kam energy.

ExampleHydrogen ke liye yeh $13.6 e V$ hai.

Upar wale dande, $n \geq 2$ , excited states (उत्तेजित अवस्थाएँ) hain. Discharge tube mein ek collision, ya theek sahi energy wala absorb hua photon, atom ko unmein se kisi par utha sakta hai. Pehla step sabse bada hai, $n = 1$ se $n = 2$ tak $10.2 e V$ ; $n = 2$ se electron ko free karne ke liye aur $3.40 e V$ kaafi hai. Ooche dande par electron bas dheele se pakda hota hai.

4/4 $n = 3$ se $n = 2$ par girte hue atom $1.89 e V$ ka photon emit karta hai.

Fig. 12.11Hydrogen ke energy levels,

E_{n} = - 13.6 e V / n^{2}

. Levels

E = 0

ki taraf paas-paas aate jaate hain; uske upar electron free hai aur uski koi bhi energy ho sakti hai.NCERT Fig. 12.7 · p. 300

Vertical energy scale par horizontal lines, left mein quantum number n aur right mein electronvolts mein energy: neeche n equals 1 minus 13.6 par, n equals 2 minus 3.40 par, n equals 3 minus 1.51 par, aur zero ke theek neeche aur lines paas-paas, zero ke upar ek hatched band jis par free likha hai. n equals 1 line par ek blue electron hai. Step 1: sirf levels. Step 2: green arrows n equals 1 se n equals 2 tak jaate hain, label 10.2 eV, aur n equals 3 tak, label 12.09 eV, aur electron n equals 3 par chadhta hai. Step 3: ek green arrow n equals 1 se seedha zero tak jaata hai, label 13.6 eV, aur electron free band mein chadh jaata hai. Step 4: ek dashed vermillion arrow n equals 3 se n equals 2 tak neeche aata hai, label 1.89 eV, aur electron n equals 2 par hai.

Diagram do aur features dikhata hai. $n$ badhne par levels paas-paas aate jaate hain, kyunki $1 / n^{2}$ kam aur kam badalta hai, aur woh $E = 0$ par mil jaate hain, jo limit $n \to \infty$ hai: infinitely door rest mein ek electron. $E = 0$ ke upar electron free hai aur uski koi bhi energy ho sakti hai, isliye wahan allowed energies alag lines ke bajaye ek continuous band banati hain. Aur excited atom zyada der excited nahi rehta. Aam taur par kuch nanoseconds mein woh neeche ke levels par gir jaata hai, har step par ek photon dete hue.

InteractiveEnergy-level lab

Hydrogen, singly ionised helium ya doubly ionised lithium ka energy-level diagram, n equals 1 se 7 tak levels aur zero ke upar ek free band ke saath. Starting level aur final level chunne par ek arrow draw hota hai, emission ke liye neeche ya absorption ke liye upar, aur photon ki electronvolts mein energy, uski frequency aur wavelength dikhti hai, saath mein ek message jo spectrum ka region aur, hydrogen ke liye, series batata hai. Emitted lines neeche ek spectrum strip mein judti hain, 10 se 5000 nanometres tak logarithmic wavelength scale par, visible band shaded. Ek slider bombarding electron beam ki energy set karta hai; Show every line button woh sabse ooncha level dhoondhta hai jahan tak beam pahunch sakti hai aur woh har line draw karta hai jo atoms phir emit kar sakte hain.

Fig. 12.12Do levels chuno aur dekho ki jump kaunsa photon emit karta hai ya use kaunsa photon chahiye. Phir gas ko electron beam se excite karo aur dekho kaun si lines dikhti hain.

Quick check

3 mein se 1

Jo hydrogen atom pehle se

n = 3

state mein hai, use ionise karne ke liye kitni energy chahiye?

Answer dekho

Answer 1.51 eV

Electron ko

E_{3} = - 13.6 / 9 = - 1.51 e V

0

tak jaana hai, isliye

1.51 e V

chahiye, ground state wale

13.6 e V

se kahin kam.

12.5 Hydrogen Atom ke Line Spectra

Bohr ka teesra postulate energy-level diagram ko spectrum mein badal deta hai. Jab electron upper state $n_{i}$ se lower state $n_{f}$ par girta hai, atom ek photon emit karta hai jiski frequency $ν_{i f}$ yeh hai

h ν_{i f} = E_{n_{i}} - E_{n_{f}}

(12.11)

Levels ek discrete set banate hain, isliye unke beech ke gaps bhi, aur har gap ek frequency fix karta hai. Isliye hydrogen spectrum sirf lines ka set ho sakta hai. (12.8) ko (12.11) mein daal kar aur $ν = c / λ$ use karke har line ki wavelength ek saath mil jaati hai:

Hydrogen ki lines ki wavelengths

\frac{1}{λ} = R (\frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}}), R = \frac{m e^{4}}{8 ε_{0}^{2} h^{3} c} = 1.097 \times 10^{7} m^{- 1}

jahan

$R$: Rydberg constant
$n_{f}, n_{i}$: final aur initial principal quantum numbers, $n_{i} > n_{f}$

Convert karne ka jaldi wala tareeka hai $λ = h c / Δ E$ , jahan $h c = 1240 e V n m$ : $Δ E$ electronvolts wale photon ki wavelength $1240 / Δ E$ nanometres hoti hai.

Jo lines same lower level par khatam hoti hain, woh ek family banati hain, jise series kehte hain. $n = 2$ tak neeche ke jumps hydrogen ki visible lines dete hain, theek wahi jinhe Balmer ne 1885 mein fit kiya tha (Fig. 12.13). Bohr ki theory ne Balmer ka formula reproduce kiya, constant $R$ ko $m$ , $e$ , $h$ aur $ε_{0}$ se nikaal kar, aur measured wavelengths se $0.1$ % se bhi behtar match kiya.

4/4 $E = 0$ se jump $3.40 e V$ deta: $365 n m$ par series limit, ultraviolet mein.

Fig. 12.13Hydrogen ki Balmer series: har jump jo

n = 2

par khatam hota hai. Lambe jumps chhoti wavelengths dete hain, aur lines

365 n m

par series limit ki taraf paas-paas aati hain.NCERT Fig. 12.5 · p. 297

Upar wala hissa energy levels n equals 2 se 8 tak dikhata hai, jo upar ki taraf paas-paas aate hain, aur zero energy par ek dashed line hai. Neeche wala hissa 350 se 700 nanometres tak ek wavelength strip hai. Har jump n equals 2 line tak neeche ek black arrow hai, jo apni banayi line ke theek upar hai, aur strip tak ek dotted guide hai. Step 1: n equals 3 se jump 656 nanometres par red line deta hai. Step 2: n equals 4 se jump 486 par sky-blue line jodta hai. Step 3: n equals 5 aur 6 se jumps 434 par blue line aur 410 par violet line jodte hain. Step 4: zero energy se ek dashed arrow 365 nanometres par series limit mark karta hai, jahan lines jhund mein aa jaati hain.

Table 12.3Hydrogen lines ki teen series, unhe dhoondhne wale physicists ke naam par

	Kahan khatam	Wavelengths (nm)
Lyman (ultraviolet)	$n_{f} = 1$	121.6 se 91.2 tak
Balmer (visible)	$n_{f} = 2$	656.3 se 364.7 tak
Paschen (infrared)	$n_{f} = 3$	1875 se 820 tak

Example

Yeh line kis jump se bani?

Hydrogen lamp $486 n m$ par ek sky-blue line dikhata hai. (a) Photon energy kya hai? (b) Kaunsa jump ise banata hai? (c) Iski frequency kya hai?

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

Wahi diagram absorption bhi samjhata hai. Ground state mein hydrogen atom photon ko tabhi le sakta hai jab photon ki energy kisi higher level tak ke gap se match kare: $10.2 e V$ , $12.09 e V$ , $12.75 e V$ waghera, ya $13.6 e V$ se upar koi bhi amount, jo use ionise kar deta hai. Continuous light ko thandi hydrogen se bhejo to woh wavelengths gayab ho jaati hain, aur Section 12.3 wali dark lines bachti hain. Kyunki thandi hydrogen ground state mein hoti hai, uski absorption lines ultraviolet Lyman lines hain. Visible Balmer wavelengths sirf woh atoms absorb karte hain jo pehle se $n = 2$ mein hain, jaise stars ki garam outer layers wale.

Nature ke kuch constants se hi Bohr ke model ne hydrogen spectrum ki har line ko samjha diya, aur isse unhe 1922 ka Physics Nobel Prize mila. Iski success ne atom ki quantum theory ko shuru karne mein bada role nibhaya.

Quick check

4 mein se 1

Jab hydrogen ka electron

n = 5

n = 2

par girta hai, to emitted photon ki wavelength nikaalo.

Answer dekho

Answer 434 nm

Δ E = 13.6 (1 / 4 - 1 / 25) = 2.86 e V

, isliye

λ = 1240 / 2.86 = 434 n m

, ek blue Balmer line.

12.6 Bohr ke Quantisation wale Second Postulate ki De Broglie Explanation

Bohr ka doosra rule sahi answers deta tha, lekin hawa se utha hua rule lagta tha. 1923 mein Louis de Broglie ne dikhaya ki yeh kahan se aata hai, pichhle chapter, Dual Nature of Radiation and Matter, wale idea se: chalte electron ki wavelength $λ = h / p$ hoti hai, jise electron-diffraction experiments ne 1927 mein confirm kiya.

Ek circle ke around wave draw karke dekho. Kahin se bhi shuru karo aur ek baar poora ghoom kar trace karo. Jab tum wapas start par pahuncho, to wave ko theek waise hi aage chalna chahiye jaise shuru hui thi, same height par aur same tarah upar jaate hue; warna doosra chakkar pehle se out of step ho jaata hai, teesra doosre se, aur chakkar dar chakkar mismatched waves ek doosre ko cancel kar deti hain. Isliye jo wave orbit ke around tikti hai, use usmein whole number baar fit hona chahiye. Yeh Class 11 mein dekhi stretched string par standing waves ka circular version hai (Fig. 12.14).

1/3Teen wavelengths theek fit hoti hain: crest crest se milta hai, aur standing wave banti hai.

Fig. 12.14Orbit ke around sirf whole number wavelengths hi standing wave bana sakti hain. (Saaf dikhane ke liye har jagah ek hi orbit use ki gayi hai; hydrogen mein

n

ki har value ki apni orbit aur apni wavelength hoti hai.)NCERT Fig. 12.8 · p. 301

Radius r ki dashed circular orbit ke centre mein ek proton hai, aur ek blue wave circle ke around lipti hai, uske aas-paas upar-neeche hoti hui. Step 1: circle ke around theek teen wavelengths fit hoti hain aur wave smoothly khud se jud jaati hai. Step 2: saadhe teen wavelengths try ki jaati hain; wave, ab dashed vermillion, apne hi start se galat direction mein milti hai, aur ek circle takkar ko mark karta hai, label ends clash. Step 3: theek paanch wavelengths fit hoti hain aur wave phir smoothly judti hai.

DerivationStanding waves se Bohr ke rule tak

Radius $r_{n}$ wali $n$ th orbit ke liye fitting condition hai

2 π r_{n} = n λ, n = 1, 2, 3, \dots

(12.12)

Light se bahut dheeme chalte electron ke liye $p = m v_{n}$ , isliye $λ = h / m v_{n}$ . Substitute karne par,

2 π r_{n} = \frac{n h}{m v_{n}} ⟹ m v_{n} r_{n} = \frac{n h}{2 π}

Yeh theek Bohr ka second postulate (12.5) hai. Quantised orbits, aur unke saath energy levels, isliye aate hain kyunki electron wave ki tarah behave karta hai, aur sirf fit hone wali standing waves hi tik sakti hain.

Example

n = 2 orbit mein electron ki wave

Hydrogen ki $n = 2$ orbit mein electron ki de Broglie wavelength nikaalo (a) uski speed se aur (b) orbit ke size se. Check karo ki dono answers match karte hain.

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

Bohr ka model, de Broglie ki picture ke support ke saath, hydrogen ke main features sahi batata hai: uska size, uski ionisation energy, aur woh frequencies jo woh emit aur absorb karta hai. Yeh kisi bhi hydrogenic atom ke liye utna hi achha kaam karta hai, yaani charge $+ Z e$ wala nucleus aur ek hi electron, jaise $H e^{+}$ ya $L i^{2 +}$ . Lekin iski serious limits hain:

Doosra electron aate hi yeh fail ho jaata hai: helium hi ise hara deta hai. Model mein sirf nucleus ka kheenchav shaamil hai. Real atom mein har electron baaki har electron ko repel bhi karta hai, aur yeh forces nuclear attraction ke barabar hoti hain, kyunki charges aur distances milte-julte hain. Iske ulat solar system mein planets ka ek doosre par kheenchav Sun ke muqable bahut chhota hai, kyunki Sun saare planets ko milakar bhi unse lagbhag 750 guna bhaari hai, isliye wahan planetary picture tik jaati hai.
Yeh intensities predict nahi kar sakta. Kuch hydrogen lines bright hain aur kuch halki, isliye kuch jumps doosron se kahin zyada baar hote honge. Bohr ke rules lines ki frequencies dete hain lekin yeh nahi batate ki har jump kitna likely hai.
Iski sharp orbits real nahi hain. Uncertainty principle ke hisaab se electron ki position aur momentum dono exactly nahi jaane ja sakte, isliye woh koi definite circular path follow nahi kar sakta.

Quantum mechanics, jo 1925 se Werner Heisenberg, Erwin Schrödinger aur doosron ne develop ki, ne orbits ki jagah probability ke clouds rakhe aur in sab sawaalon ke jawaab diye. To phir galat maane gaye model ko kyun seekhein? Kyunki sirf Newton, Coulomb aur ek quantum rule se yeh hydrogen ki energies bilkul sahi deta hai; kyunki iske numbers ( $a_{0} = 0.053 n m$ , $13.6 e V$ , $2.2 \times 10^{6} m / s$ ) aaj bhi atomic physics ka scale set karte hain; aur kyunki yeh method ka ek sabak hai. Bohr ne ek objection side mein rakha jiska woh jawaab nahi de sakte the, sharp predictions kiye, aur experiment ko faisla karne diya.

Definition

Hydrogenic atom

Charge $+ Z e$ wale nucleus aur ek hi electron wala atom ya ion, jismein koi electron–electron forces nahi hote.

Example $H$ , $H e^{+}$ aur $L i^{2 +}$ .

Quick check

3 mein se 1

De Broglie ke hisaab se hydrogen ki

n = 3

orbit ke around kitni electron wavelengths fit hoti hain?

1369

Answer dekho

Answer (B) 3

(12.12) dekho.

Flash cards · 10

Formula cards

1 / 10

Recap

Summary aur key points

Atom neutral hai: uske halke electrons ko barabar positive charge balance karta hai jismein lagbhag saara mass hota hai. Thomson ne us charge ko poore atom mein phailaya (Section 12.1).
Geiger aur Marsden ne paaya ki kuch α-particles thin gold foil se wapas uchhalte hain. Aisa sirf ek chhota, bhaari, positive nucleus kar sakta hai, jo atom se $10^{4}$ se $10^{5}$ guna chhota hai (Section 12.2).
Chhota impact parameter $b$ bada scattering angle deta hai; head-on α-particle $d = 2 Z e^{2} / 4 π ε_{0} K$ par rukta hai aur wapas mudta hai (Section 12.2.1).
Circular orbit mein electron ke liye $E = - e^{2} / 8 π ε_{0} r$ , saath mein $K = - E$ aur $U = 2 E$ ; negative energy ka matlab hai ki woh bound hai (Section 12.2.2).
Thin chamakti gas emission line spectrum deti hai jo element ko pehchanta hai; thandi gas white light se wahi wavelengths absorb karti hai (Section 12.3).
Classically, orbit mein ghoomta electron radiate karta hai, lagbhag $10^{- 11} s$ mein spiral karke nucleus mein gir jaata hai aur continuous spectrum emit karta hai, isliye Rutherford ka atom poori kahani nahi ho sakta (Section 12.4).
Bohr ke postulates: stationary orbits radiate nahi karti; $L = n h / 2 π$ ; ek jump $h ν = E_{i} - E_{f}$ wala ek photon emit karta hai (postulates).
Hydrogen ke liye $r_{n} = n^{2} a_{0}$ , jahan $a_{0} = 0.0529 n m$ , $v_{n} \propto 1 / n$ aur $E_{n} = - 13.6 e V / n^{2}$ (derivation).
Ground state $n = 1$ hai; ionisation energy $13.6 e V$ hai; excited states $E = 0$ ki taraf paas-paas aati hain, jiske upar electron free hai (Section 12.4.1).
Har spectral line ek jump hai, $h ν_{i f} = E_{n_{i}} - E_{n_{f}}$ ; $n = 1$ , $2$ , $3$ par khatam hone wale jumps Lyman (ultraviolet), Balmer (visible) aur Paschen (infrared) series dete hain (Section 12.5).
De Broglie ne quantisation rule samjhaya: orbit ke around electron wavelengths ki whole number fit honi chahiye, $2 π r_{n} = n λ$ (Section 12.6).
Bohr ka model sirf one-electron (hydrogenic) atoms ke liye kaam karta hai aur line intensities predict nahi kar sakta; quantum mechanics ne iski jagah li (limitations).

Practice · 14 questions

Practice questions answers ke saath (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises 12.1
Har observation ya prediction ke liye tay karo ki woh atom ke kis model ka hai: Thomson ka, Rutherford ka, dono ka, ya kisi ka nahi.
1. Gold foil ke peeche $150^{\circ}$ par detector ke saath har minute kuch flashes dikhti hain.
  Thomson ka modelRutherford ka modelDono modelsKoi model nahi
  Answer dekho
  Answer (B) Rutherford ka model
  α-particle ko wapas mudne ke liye concentrated nucleus ke paas wala intense field chahiye. Phaila hua Thomson atom $150^{\circ}$ par lagbhag kuch bhi nahi bhejta.
2. Atom ke centre mein rest mein rakha electron wahin rehta hai, aur halka sa dhakka dene par centre ke around oscillate karta hai.
  Thomson ka modelRutherford ka modelDono modelsKoi model nahi
  Answer dekho
  Answer (A) Thomson ka model
  Positive charge ki uniform ball ke andar field wapas centre ki taraf point karta hai aur centre se distance ke saath badhta hai, isliye wahan electron stable equilibrium mein hai. Nuclear atom mein rest wala electron seedha nucleus mein gir jaata.
3. Atom ke andar density har jagah lagbhag solid gold jitni hai, lagbhag $2 \times 10^{4} k g / m^{3}$ .
  Thomson ka modelRutherford ka modelDono modelsKoi model nahi
  Answer dekho
  Answer (A) Thomson ka model
  Thomson ka atom apna mass poore volume mein phailata hai, isliye uski density bas metal ki bulk density hai. Rutherford ke atom mein lagbhag saara mass nucleus mein hai jiski density lagbhag $10^{17} k g / m^{3}$ hai, jabki baaki atom lagbhag khaali hai.
4. Classical electromagnetism predict karta hai ki atom lagbhag $10^{- 11} s$ mein collapse ho jaata hai.
  Thomson ka modelRutherford ka modelDono modelsKoi model nahi
  Answer dekho
  Answer (B) Rutherford ka model
  Sirf orbit mein ghoomte, aur isliye accelerate hote, electrons hi apni energy radiate karke khote hain. Yahi nuclear model ki kamzori hai, aur Bohr ke postulates ise theek karne ke liye banaye gaye.
5. Hydrogen sirf kuch sharp wavelengths par hi light deti hai.
  Thomson ka modelRutherford ka modelDono modelsKoi model nahi
  Answer dekho
  Answer (D) Koi model nahi
  Koi bhi classical model line spectra nahi samjha sakta. Thomson ka atom khaas wavelengths ka koi kaaran nahi deta, aur Rutherford ka atom electron ke spiral karke andar jaane par continuous sweep emit karta. Lines samjhane ke liye Bohr ke quantum rules chahiye the.
2
Long answerMediumNCERT Exercises 12.2
Ek student $5 M e V$ α-particles ke saath scattering experiment dohraata hai, lekin gold ki jagah ek thin polythene film use karta hai, jismein sirf carbon nuclei ( $Z = 6$ , mass $12 u$ ) aur hydrogen nuclei ( $Z = 1$ , mass $1 u$ ) hain. α-particle ka mass $4 u$ hai.
1. Kya hydrogen nucleus (proton) α-particle ko wapas source ki taraf mod sakta hai?
  Haan, head-on collision meinHaan, lekin sirf dheeme α-particles ke liyeNahi: proton α-particle se halka hai aur use ghuma kar wapas nahi bhej sakta
  Answer dekho
  Answer (C) Nahi: proton α-particle se halka hai aur use ghuma kar wapas nahi bhej sakta
  Bhaari particle jab rest mein pade halke particle se takraata hai, to hamesha aage hi chalta rehta hai, jaise cricket ball table-tennis ball ko mushkil se mehsoos karti hai. Momentum aur energy conservation uske deflection ko $\sin θ_{\max} = m_{p} / m_{α} \approx 1 / 4$ , yaani lagbhag $15^{\circ}$ , tak limit karte hain; iske bajaye proton aage dhakel diya jaata hai.
2. Gold ke muqable carbon nuclei se large-angle scattering kitna common hoga?
  Zyada common, kyunki carbon nuclei chhote hainLagbhag utna hi common, kyunki carbon nuclei bhi α-particle se bhaari hainKahin zyada rare, kyunki carbon nucleus par bahut kam charge hai
  Answer dekho
  Answer (C) Kahin zyada rare, kyunki carbon nucleus par bahut kam charge hai
  Carbon α-particle se bhaari hai, isliye woh use wapas mod sakta hai, lekin uska charge $6 e$ gold ke $79 e$ se lagbhag 13 guna chhota hai. Uska field bhi utna hi kamzor hai, isliye strongly deflect hone ke liye α-particle ko kahin zyada paas se guzarna padega, aur aise close passes kahin zyada rare hain.
3. Experiment film ki hydrogen ke baare mein kya dikhayega, aur kya nahi dikha payega?
  Solution dikhao
  Yeh film se aage nikalte protons dikhayega, jinhe bahut paas se guzre α-particles ne dhakela. Jis proton ko itni zor se maara ja sake, woh ek chhota, bhaari, charged object hona chahiye, isliye hydrogen ka bhi nucleus hai.
  Yeh hydrogen se back-scattering nahi dikha sakta, aur sirf halke, kam charge wale nuclei ke saath gold wala dramatic large-angle evidence zyaadatar gayab hoga. Gold achha choice tha kyunki uska nucleus bhaari bhi hai aur highly charged bhi, aur kyunki use peet kar bahut patla leaf banaya ja sakta hai.
3
NumericalEasyNCERT Exercises 12.3
Sodium atom ke do energy levels $2.10 e V$ ke gap par hain. Atom upper se lower level par girta hai.
1. Emitted light ki frequency nikaalo.
  Hz
  Answer dekho
  Answer 508000000000000 Hz
  $ν = Δ E / h = (2.10 \times 1.602 \times 10^{- 19} J) / (6.626 \times 10^{- 34} J s) = 5.08 \times 10^{14} H z$ .
2. Uski wavelength nikaalo.
  nm
  Answer dekho
  Answer 590 nm
  $λ = 1240 / 2.10 = 590 n m$ , ya $c / ν = (3.00 \times 10^{8}) / (5.08 \times 10^{14}) = 5.90 \times 10^{- 7} m$ .
3. Iska colour kya hai?
  Solution dikhao
  AnswerYellow.
  $590 n m$ visible spectrum ke yellow part mein hai. Yeh sodium street lamps ki jaani-pehchaani roshni hai.
4
NumericalEasyNCERT Exercises 12.4
Hydrogen ke electron ki first excited state ( $n = 2$ ) mein total energy $- 3.40 e V$ hai.
1. Electron ki kinetic energy kya hai?
  eV
  Answer dekho
  Answer 3.4 eV
  $K = - E = + 3.40 e V$ .
2. Uski potential energy kya hai?
  eV
  Answer dekho
  Answer -6.8 eV
  $U = 2 E = - 6.80 e V$ . Check: $K + U = 3.40 - 6.80 = - 3.40 e V$ .
5
NumericalMediumNCERT Exercises 12.5
$n = 2$ state mein ek hydrogen atom photon absorb karta hai aur $n = 5$ state mein pahunch jaata hai.
1. Photon ki energy nikaalo.
  eV
  Answer dekho
  Answer 2.856 eV
  $E_{5} - E_{2} = - 0.544 - (- 3.40) = 2.86 e V$ .
2. Uski wavelength nikaalo.
  nm
  Answer dekho
  Answer 434 nm
  $λ = 1240 / 2.856 = 434 n m$ (blue).
3. Uski frequency nikaalo.
  Hz
  Answer dekho
  Answer 691000000000000 Hz
  $ν = (2.856 \times 1.602 \times 10^{- 19} J) / (6.626 \times 10^{- 34} J s) = 6.91 \times 10^{14} H z$ .
6
NumericalMediumNCERT Exercises 12.6
Hydrogen ka Bohr model use karo.
1. $n = 2$ aur $n = 4$ orbits mein electron ki speed nikaalo.
  Solution dikhao
  $v_{2} = 2.19 \times 10^{6} / 2 = 1.09 \times 10^{6} m / s$ aur $v_{4} = 2.19 \times 10^{6} / 4 = 5.47 \times 10^{5} m / s$ .
2. Har orbit mein ek revolution ka time nikaalo.
  Solution dikhao
  $r_{2} = 4 a_{0} = 2.12 \times 10^{- 10} m$ , isliye $T_{2} = 2 π (2.12 \times 10^{- 10}) / (1.09 \times 10^{6}) = 1.22 \times 10^{- 15} s$ .
  $r_{4} = 16 a_{0} = 8.46 \times 10^{- 10} m$ , isliye $T_{4} = 2 π (8.46 \times 10^{- 10}) / (5.47 \times 10^{5}) = 9.72 \times 10^{- 15} s$ .
  Check: $T \propto r / v \propto n^{3}$ , aur $T_{4} / T_{2} = 2^{3} = 8$ .
3. Atom aam taur par photon emit karne se pehle lagbhag $1 n s$ tak $n = 2$ state mein rehta hai. Itne time mein electron lagbhag kitne chakkar lagata hai?
  revolutions
  Answer dekho
  Answer 820000 revolutions
  $(10^{- 9} s) / (1.22 \times 10^{- 15} s) \approx 8.2 \times 10^{5}$ : lagbhag das lakh chakkar, isliye apni motion ke scale par stationary state kaafi der tak tikti hai.
7
NumericalEasyNCERT Exercises 12.7
Bohr radius $a_{0} = 5.29 \times 10^{- 11} m$ hai.
1. $n = 5$ orbit ka radius kya hai?
  m
  Answer dekho
  Answer 1.32e-9 m
  $r_{5} = 25 \times 5.29 \times 10^{- 11} = 1.32 \times 10^{- 9} m$ .
2. Physicists hydrogen atoms ko bahut badi 'Rydberg' states mein excite kar sakte hain. Kaunsa $n$ radius $1.32 \times 10^{- 7} m$ ki orbit deta hai, jo ek chhote bacterium ke size jitna hai?
  Answer dekho
  Answer 50
  $n^{2} = (1.32 \times 10^{- 7}) / (5.29 \times 10^{- 11}) = 2495$ , isliye $n = 49.95 \approx 50$ .
8
Long answerMediumNCERT Exercises 12.8
Discharge tube mein electrons ko $13.0 V$ se accelerate kiya jaata hai aur phir woh room temperature par hydrogen atoms se takraate hain.
1. Beam ground-state atoms ko kis level tak excite kar sakti hai?
  Answer dekho
  Answer 4
  $E_{4} - E_{1} = 12.75 e V \leq 13.0 e V$ , lekin $E_{5} - E_{1} = 13.06 e V$ zyada hai. Isliye $n = 4$ .
2. Ground state mein wapas aate hue excited atoms jo bhi lines emit kar sakte hain, sabki list unki wavelength ke saath banao.
  Solution dikhao
  $n = 4$ se electron kisi bhi raaste neeche aa sakta hai, isliye levels 1 se 4 ke beech har jump hota hai: $4 \times 3 / 2 = 6$ lines.
  Table 12.4
  Jump $Δ E$ (eV) $λ$ (nm) Series
  4 → 1 12.75 97.2 Lyman
  3 → 1 12.09 102.6 Lyman
  2 → 1 10.20 121.6 Lyman
  4 → 2 2.55 486.2 Balmer
  3 → 2 1.89 656.4 Balmer
  4 → 3 0.661 1875 Paschen
3. (c)
  Inmein se kaun si lines tum apni aankhon se dekh sakte ho?
  $97.2 n m$ $121.6 n m$ $486.2 n m$ $656.4 n m$ $1875 n m$
  Saare sahi options chuno
  Answer dekho
  Answer
  (C) $486.2 n m$
  (D) $656.4 n m$
  Sirf do Balmer lines, $486 n m$ par sky blue aur $656 n m$ par red, lagbhag $400$ se $700 n m$ ki visible range mein aati hain.
9
NumericalHardNCERT Exercises 12.9
Earth ke around Moon ki orbit ko Bohr orbit ki tarah treat karo. Moon ka mass $7.35 \times 10^{22} k g$ , uski orbit ka radius $3.84 \times 10^{8} m$ aur orbital speed $1.02 \times 10^{3} m / s$ hai.
1. Quantum number $n$ nikaalo.
  Answer dekho
  Answer 2.73e+68
  $L = m v r = (7.35 \times 10^{22}) (1.02 \times 10^{3}) (3.84 \times 10^{8}) = 2.88 \times 10^{34} k g m^{2} / s$ , isliye $n = 2 π L / h = 2 π (2.88 \times 10^{34}) / (6.63 \times 10^{- 34}) = 2.73 \times 10^{68}$ .
2. Isse rozmarra ke systems mein quantisation ke baare mein kya pata chalta hai?
  Solution dikhao
  Jab $n$ lagbhag $10^{68}$ ho, to padosi allowed orbits ka angular momentum lagbhag $10^{68}$ mein ek part se alag hota hai. Koi bhi measurement aise steps kabhi detect nahi kar sakta, isliye bade objects ke liye allowed values ek continuum banati hain aur classical mechanics bilkul sahi kaam karti hai. Quantisation tabhi matter karta hai jab $L$ , $h / 2 π$ ka kuch hi guna ho, jaise atoms mein electrons ke liye.
Quick answerEasy
$n = 4$ state wale hydrogen atom ko ionise karne ke liye kitni energy chahiye?
eV
Answer dekho
Answer 0.85 eV
$E_{4} = - 13.6 / 16 = - 0.85 e V$ , isliye $0.85 e V$ electron ko free kar deta hai.
11
Quick answerEasy
Hydrogen ki $n$ th orbit ki har quantity ko match karo ki woh $n$ par kaise depend karti hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
Orbit radius : $\propto n^{2}$
Electron speed : $\propto 1 / n$
Total energy : $\propto - 1 / n^{2}$
Ek revolution ka time : $\propto n^{3}$
$T = 2 π r / v \propto n^{2} / (1 / n) = n^{3}$ , aur $E \propto - 1 / r \propto - 1 / n^{2}$ .
12
Quick answerEasy
Rutherford ki reasoning ke steps ko order mein lagao.
1. Isliye atom zyaadatar khaali hai, ek chhote, dense, positive nucleus ke saath
2. Wapas uchhalne ke liye chhoti distance par bahut bada repulsive force chahiye
3. Zyaadatar α-particles thin foil se seedhe nikal jaate hain, lekin kuch wapas uchhalte hain
4. Aise force ke liye positive charge aur mass ek chhote region mein pack hone chahiye
Answer dekho
Answer
Zyaadatar α-particles thin foil se seedhe nikal jaate hain, lekin kuch wapas uchhalte hain
Wapas uchhalne ke liye chhoti distance par bahut bada repulsive force chahiye
Aise force ke liye positive charge aur mass ek chhote region mein pack hone chahiye
Isliye atom zyaadatar khaali hai, ek chhote, dense, positive nucleus ke saath
Observation, phir uske liye zaroori force, phir woh force kaun de sakta hai, phir model.
Quick answerMedium
Ground state mein hydrogen atom $11.0 e V$ energy ka photon absorb kar sakta hai.
SahiGalat
Answer dekho
Answer Galat
$n = 1$ se gaps $10.2$ , $12.09$ , $12.75 e V$ waghera hain, aur ionisation ke liye $13.6 e V$ chahiye. $11.0 e V$ ka photon inmein se kisi se match nahi karta, isliye woh nikal jaata hai. ( $11.0 e V$ ka takraane wala electron atom ko excite kar sakta hai, kyunki woh extra energy apne paas rakh sakta hai.)
Quick answerEasy
Hydrogen ke ground state se electron nikaalne ke liye zaroori energy, $13.6 e V$ , ko uski energy kehte hain.
Answer dekho
Answer ionisation
Electron nikaalne se ek positive ion bachta hai, isliye yeh ionisation energy hai.

Jump	$Δ E$ (eV)	$λ$ (nm)	Series
4 → 1	12.75	97.2	Lyman
3 → 1	12.09	102.6	Lyman
2 → 1	10.20	121.6	Lyman
4 → 2	2.55	486.2	Balmer
3 → 2	1.89	656.4	Balmer
4 → 3	0.661	1875	Paschen

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 12.1	Practice Q1
NCERT Exercises 12.2	Practice Q2
NCERT Exercises 12.3	Practice Q3
NCERT Exercises 12.4	Practice Q4
NCERT Exercises 12.5	Practice Q5
NCERT Exercises 12.6	Practice Q6
NCERT Exercises 12.7	Practice Q7
NCERT Exercises 12.8	Practice Q8
NCERT Exercises 12.9	Practice Q9

Quick check: MCQs

Warm-up2 questions
12.2.1 Alpha-particle trajectory4 questions
12.2.2 Electron orbits3 questions
12.3 Atomic Spectra3 questions
12.4 Hydrogen Atom ka Bohr Model4 questions
12.4.1 Energy levels3 questions
12.5 Hydrogen Atom ke Line Spectra4 questions
12.6 Bohr ke Quantisation wale Second Postulate ki De Broglie Explanation3 questions

Glossary · 12

Key terms (English + Hindi)

absorption spectrumअवशोषण स्पेक्ट्रमavashoshan spectrum: Continuous spectrum jismein dark lines hain jahan thandi gas ne apni characteristic wavelengths absorb kar li hain.Chapter mein pehli baar yahan aaya →

Bohr radiusबोर त्रिज्याBohr trijya: Hydrogen ki sabse chhoti orbit ka radius, $a_{0} = 5.29 \times 10^{- 11} m$ .Chapter mein pehli baar yahan aaya →

emission line spectrumउत्सर्जन रेखीय स्पेक्ट्रमutsarjan rekhiya spectrum: Low pressure par kisi element ki excited gas se, us element ki characteristic wavelengths par bright lines.Chapter mein pehli baar yahan aaya →
excited stateउत्तेजित अवस्थाuttejit avastha: Ground state ke upar atom ki koi bhi state (hydrogen mein $n \geq 2$ ).Chapter mein pehli baar yahan aaya →

ground stateनिम्नतम अवस्थाnimnatam avastha: Atom ki sabse kam energy wali state; hydrogen ke liye $n = 1$ , $E_{1} = - 13.6 e V$ .Chapter mein pehli baar yahan aaya →

hydrogenic atomहाइड्रोजन-सम परमाणुhydrogen-sam paramanu: Charge $+ Z e$ wala nucleus aur ek hi electron, jaise $H e^{+}$ ya $L i^{2 +}$ .Chapter mein pehli baar yahan aaya →

impact parameterसंघट्ट प्राचलsanghatt prachal: Particle ki initial line of motion aur target nucleus ke centre ke beech ki perpendicular distance.Chapter mein pehli baar yahan aaya →
ionisation energyआयनन ऊर्जाaayanan oorja: Ground state wale atom se electron ko free karne wali sabse kam energy; hydrogen ke liye $13.6 e V$ .Chapter mein pehli baar yahan aaya →

nucleusनाभिकnabhik: Atom ka chhota, dense, positive centre, jismein uska saara positive charge aur lagbhag saara mass hota hai.Chapter mein pehli baar yahan aaya →

principal quantum numberमुख्य क्वांटम संख्याmukhya quantum sankhya: Bohr orbit ko label karne wala whole number $n$ ; $L = n h / 2 π$ aur $E_{n} = - 13.6 e V / n^{2}$ .Chapter mein pehli baar yahan aaya →

stationary stateस्थायी अवस्थाsthayi avastha: Atom ki allowed state jismein electron radiate nahi karta aur energy constant rehti hai.Chapter mein pehli baar yahan aaya →

α-particleऐल्फ़ा कणalpha kan: Helium nucleus (do protons, do neutrons), charge $+ 2 e$ , jo kuch radioactive nuclei nikaalte hain.Chapter mein pehli baar yahan aaya →

12.1 Introduction

12.2 Alpha-particle Scattering aur Rutherford ka Nuclear Model of Atom

12.2.1 Alpha-particle trajectory

12.2.2 Electron orbits

12.3 Atomic Spectra

12.4 Hydrogen Atom ka Bohr Model

12.4.1 Energy levels

12.5 Hydrogen Atom ke Line Spectra

12.6 Bohr ke Quantisation wale Second Postulate ki De Broglie Explanation

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye