Dual Nature of Radiation and Matter: Class 12 Physics Chapter 11 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

85 min
15 practice questions
29 MCQs

Is chapter mein aap kya seekhenge

Samjho ki cathode-ray experiments ne electron ko kaise dhoondha aur uska specific charge, charge aur mass kaise naapa
Work function aur electron volt ko define karo, aur thermionic, field aur photoelectric emission ko compare karo
Hertz, Hallwachs aur Lenard ke observations samjho, aur photoelectric emission study karne wala apparatus bhi
Photocurrent vs intensity, photocurrent vs collector potential, aur stopping potential vs frequency ke graphs ko interpret karo
Samjho ki photoelectric effect ke liye light ki wave picture kyun fail hoti hai aur Einstein ke light quanta kaise kaam karte hain
Einstein ki photoelectric equation se threshold frequency, stopping potential, maximum kinetic energy aur Planck's constant nikaalo
Photons ki properties batao, aur photon ki energy, momentum aur ek beam mein per second photons ki sankhya calculate karo
Particles par de Broglie relation lagao aur samjho ki matter waves sirf atomic scale par hi kyun dikhti hain

NCERT Physics Class 12, Chapter 11, pp. 274–289 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

11.1 Introduction

Ab tak light ke wave hone ka case pakka lagta hai. Do slits bright aur dark fringes dete hain, ek patla gap beam ko phaila deta hai, aur Polaroid reflection ki glare kaat sakta hai, jaise tumne Wave Optics mein dekha. Maxwell's equations to wave ki speed bhi predict kar deti hain. Phir bhi camera ka light meter, particle detector mein single flashes ginne wala photomultiplier tube, aur night-vision tube, sab ek aise effect par chalte hain jise wave picture explain nahi kar sakti. Yeh chapter us puzzle ko uske answer tak follow karta hai, aur phir answer ko ulta ghuma deta hai: agar light particles ki stream jaisa behave kar sakti hai, to kya particles waves jaisa behave kar sakte hain?

Yeh puzzle unnisvi sadi ke aakhri chauthai hisse ke experiments se nikla. 1887 mein Heinrich Hertz ne electromagnetic waves banayi aur detect ki, jo Maxwell ki wave theory ki badi jeet thi. Usi samay physicists bahut kam pressure wali gases se electricity guzaar rahe the, aur yeh experiments surprises dete ja rahe the: X-rays (Wilhelm Röntgen, 1895) aur, sabse badhkar, electron.

Ek glass tube mein do metal electrodes seal karo, lagbhag saari hawa pump karke nikaal do aur unke across kuch hazaar volts lagao. Jab pressure mercury ke lagbhag ek millimetre ke hazaarven hisse tak, yaani atmosphere ke lagbhag ek millionth tak gir jaata hai, to negative electrode se door wale end par glass chamakne lagta hai. Iska colour glass par depend karta hai; ordinary soda glass green chamakta hai. Cathode se kuch nikal kar glass se takra raha hai. In cathode rays (कैथोड किरणें) ko William Crookes ne 1870s mein dhyaan se study kiya, aur 1879 mein unhone kaha ki yeh tez, negatively charged particles ki streams hain.

1897 mein J. J. Thomson ne is idea ko test kiya: cathode rays ka ek beam, do plates ke beech ek electric field, aur ek magnetic field jo beam aur electric field dono ke right angle par tha (Fig. 11.1). Har field akele beam ko mod deta hai; dono ko saath lagakar adjust karo to woh ek doosre ko cancel kar sakte hain. Is cancellation se particles ki speed milti hai, aur phir sirf ek field se hone wali bending se unka charge-to-mass ratio, yaani specific charge $e / m$ milta hai. Pata chala ki particles light ki speed ke lagbhag dasven se paanchven hisse tak ki speed se chalte hain, $c = 3 \times 10^{8} m / s$ .

3/3Page ke andar jaata magnetic field unhe neeche dhakelta hai. Jab $e E = e v B$ hota hai to beam phir seedha ho jaata hai, isliye $v = E / B$ .

Fig. 11.1Beam par electric aur magnetic force ko balance karke particles ki speed naapi jaati hai.

Figure ke across ek lamba evacuated tube hai: left end par cathode, phir slit wala anode, horizontal plates ka ek pair (upar positive, neeche negative) aur right end par screen. Step 1: electron beam cathode se seedha screen ke centre par ek chamakte spot tak jaata hai. Step 2: plates charge hone par beam positive plate ki taraf upar mudta hai aur spot upar chala jaata hai; ek arrow upward electric force dikhata hai. Step 3: plates ke upar page ke andar jaata ek magnetic field lagaya jaata hai, jo crossed-circle symbols se dikhaya gaya hai; ek doosra, neeche ki taraf magnetic force electric force ko balance karta hai aur beam phir seedha chalta hai, jo tab hota hai jab v, E divided by B ke barabar ho.

Do findings ne is particle ko khaas bana diya. Iska $e / m$ same aaya, chahe cathode kisi bhi metal ka ho aur tube mein koi bhi gas bachi ho. Aur ultraviolet light se metal se nikaale gaye, ya white-hot filament se ubaal kar nikaale gaye negative particles ka bhi bilkul yahi $e / m$ tha. Ek hi tarah ka particle, bilkul alag materials se bilkul alag tareekon se bana, zaroor saare matter ka common ingredient hoga. Thomson ne ise electron kaha; iske specific charge ki aaj ki accepted value $e / m = 1.76 \times 10^{11} C / k g$ hai. Unhe 1906 mein Physics ka Nobel Prize mila.

Khud charge R. A. Millikan ke oil-drop experiment se mila, jo 1913 mein publish hua. Millikan ne do charged plates ke beech upar-neeche hoti chhoti charged oil drops ka time naapa, aur paaya ki har drop par charge ek sabse chhote charge, $e = 1.602 \times 10^{- 19} C$ , ka whole-number multiple tha. Charge quantised hai, jaisa tumne Part I ke Chapter 1 mein maana tha. Phir $e$ ko $e / m$ se divide karke electron ka mass mila, lagbhag $9.11 \times 10^{- 31} k g$ , jo sabse halke atom se bhi lagbhag do hazaar guna halka hai.

Example

Do fields se electron ko tolna

Ek cathode-ray tube mein deflecting plates $1.5 c m$ door hain aur unke across $450 V$ hai. $0.75 m T$ ka magnetic field, jo beam aur electric field dono ke perpendicular hai, beam ko seedha chala deta hai. Electric field off karne par wahi magnetic field beam ko $30.3 c m$ radius ke arc mein mod deta hai. $e = 1.602 \times 10^{- 19} C$ lekar nikaalo (a) electrons ki speed, (b) unka specific charge aur (c) unka mass.

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

Timeline

Cathode rays se matter waves tak

1879
Crookes: cathode rays charged particles hain
1887
Hertz: ultraviolet light se sparks aasaani se jump karte hain
1888
Hallwachs: ultraviolet light zinc ko discharge karti hai
1897
Thomson ne $e / m$ naapa: electron
1900
Planck: radiation energy $h ν$ ke steps mein
Garam bodies se nikalne wali radiation ka spectrum explain karne ke liye laaya gaya.
1902
Lenard: electron energy intensity ko ignore karti hai
1905
Einstein: light quanta photoemission explain karte hain
1913
Millikan ki oil drops se $e$ fix hua
1916
Millikan ne Einstein ki equation confirm ki
1923
Compton: X-ray photons electrons se takra kar uchhalte hain
1924
de Broglie: particles ki bhi waves hoti hain
1927
Electron diffraction dekha gaya
USA mein C. J. Davisson aur L. H. Germer; Britain mein G. P. Thomson.

To 1900s ki shuruaat tak electron ek definite charge aur mass wala particle tha, aur light ek wave thi. Is chapter ka baaki hissa dikhata hai ki in dono baaton mein se har ek sirf aadhi kahani hai: radiation aur matter dono ki dual nature (द्वैत प्रकृति) hoti hai.

Quick check

3 mein se 1

Ek cathode-ray beam crossed electric aur magnetic fields se bina deflect hue guzarta hai. Isse particles ke baare mein kya pata chalta hai?

Unka specific charge

E / B

ke barabar haiUn par koi electric charge nahi haiUnki speed

E / B

ke barabar haiUnka mass

e B / E

ke barabar hai

Answer dekho

Answer (C) Unki speed $E / B$ ke barabar hai

Dono forces barabar aur opposite hain,

e E = e v B

, isliye

v = E / B

e / m

nikaalne ke liye ek doosra measurement chahiye, jaise sirf magnetic field mein path ka radius. Uncharged beam bhi seedha jaata, lekin woh akele kisi bhi field se mudta nahi.

11.2 Electron emission

Copper wire free electrons se bhari hoti hai; isiliye woh itna achha conduct karti hai. Phir bhi table par padi wire kamre mein electrons nahi chhidakti. Metal ke andar gehraai mein, har free electron ke har taraf positive ions hote hain aur unke khinchaav cancel ho jaate hain. Surface par yeh balance toot jaata hai. Jo electron bahar bhatak jaata hai, woh metal ko thoda positive chhod deta hai, aur peeche chhoote ions ka attraction use waapas kheench leta hai.

Hamesha ke liye nikalne ke liye electron ko ek certain minimum energy deni padti hai. Metal surface se electron ko free karne ke liye zaroori yeh sabse kam energy metal ka work function (कार्य फलन) kehlati hai, aur ise $ϕ_{0}$ likhte hain. Iski value metal par aur surface ki condition par bhi depend karti hai: ek patli oxide layer, adsorbed gas ki film ya kisi doosre metal ka zara sa trace ise kaafi badal sakta hai.

Definition

Work function

Woh minimum energy $ϕ_{0}$ jo kisi electron ko metal ki surface se free karne ke liye deni padti hai.

ExampleCaesium ke liye lagbhag $2.1 e V$ aur tungsten ke liye lagbhag $4.5 e V$ .

Work functions kuch guna $10^{- 19} J$ hote hain, jo saath lekar chalne mein bade awkward numbers hain. Isliye atomic aur nuclear physics ek chhota unit use karti hai, electron volt: woh kinetic energy jo ek electron rest se $1 V$ ke potential difference se accelerate hone par paata hai. Kyunki milne wali energy $q V$ hai,

1 e V = (1.602 \times 10^{- 19} C) (1 V) = 1.602 \times 10^{- 19} J

Is scale par work functions caesium ke lagbhag $2 e V$ se platinum ke lagbhag $6 e V$ tak hote hain. $1000 V$ se accelerate hua electron $1000 e V$ , yaani $1 k e V$ , paata hai, apparatus ka size chahe jo bhi ho.

Metal ke electrons ko yeh energy kaise di ja sakti hai? Teen practical tareeke hain, jinhe Fig. 11.2 mein compare kiya gaya hai.

Heating. High temperature par kuch free electrons ko itni thermal energy mil jaati hai ki woh surface paar kar lete hain. Yeh thermionic emission X-ray tubes aur electron guns mein chamakte tungsten filament se electrons deta hai.
Bahut strong electric field. Metal surface par $10^{8} V / m$ order ka field electrons ko seedha bahar kheench sakta hai. Yeh field emission hai; kuch electron microscopes mein electron source ke roop mein use hone wale bahut nukeele tips isi tarah kaam karte hain.
Light. Suitable frequency ki light metal par padne se usmein se electrons free ho sakte hain. Yeh photoelectric emission hai, aur isse free hone wale electrons ko photoelectrons (प्रकाशिक इलेक्ट्रॉन) kehte hain.

4/4Suitable frequency ki light electrons ko free karti hai: photoelectric emission.

Fig. 11.2Bahar nikalne ke liye sabse energetic electrons ko bhi energy

ϕ_{0}

deni padti hai, heat, strong field ya light se.

Left mein ek energy diagram: electrons metal ko represent karne wale ek well ke neeche wale hisse ko bharte hain, aur top par ek dashed line metal ke bahar rest par ek electron ko mark karti hai. Bhare hisse ke top aur dashed line ke beech ek double arrow par phi zero, yaani work function, likha hai. Right mein ek panel har step ke saath badalta hai. Step 1: metal block ke theek bahar ek electron ko positive ions waapas kheench lete hain. Step 2: flames metal ko garam karti hain, label thermionic emission hai, aur diagram mein ek electron well se upar nikal jaata hai. Step 3: ek positively charged plate metal ke saamne hai aur unke beech field lines hain, label field emission hai, aur electron phir nikal jaata hai. Step 4: ek wavy light ray metal se takrati hai, label photoelectric emission hai, aur electron nikal jaata hai.

Quick check

3 mein se 1

Ek electron rest se start hokar

250 V

ke potential difference se accelerate hota hai. Use kitni kinetic energy milti hai, joules mein?

Answer dekho

Answer 4.005e-17 J

Use

250 e V = 250 \times 1.602 \times 10^{- 19} J = 4.0 \times 10^{- 17} J

milti hai.

11.3 Photoelectric effect

Electrons ko free karne ke teen tareekon mein se teesre mein sabse gehra sabak chhupa tha. Ise pehli baar accident se notice kiya gaya.

11.3.1 Hertz ke observations

1887 mein Heinrich Hertz (1857–1894) apni laboratory mein electromagnetic waves bhej rahe the. Unka detector ek chhote gap wala wire ka loop tha, aur receive hui wave gap ke across ek spark ke roop mein dikhti thi. Hertz ne notice kiya ki jab ultraviolet light, transmitter ke spark se ya arc lamp se, gap ke paas metal par padti thi to detector zyada aasaani se spark karta tha. Gap ko light se dhakne par sparks kamzor ho jaate the. Ab hum jaante hain ki kya ho raha tha. Ultraviolet light kuch surface electrons ko woh energy de rahi thi jo unhe metal ke ions se alag hone ke liye chahiye thi, aur yeh free charges har spark ko shuru hone mein madad karte the.

11.3.2 Hallwachs aur Lenard ke observations

Wilhelm Hallwachs ne 1888 mein is clue ko aage badhaya, ek gold-leaf electroscope ke top par lagi saaf zinc plate ke saath (Fig. 11.3). Unhe teen cheezein mili. Negatively charged plate par ultraviolet light padne se uska charge khatam ho gaya. Uncharged plate usi light mein positively charged ho gayi. Aur jo plate pehle se positive thi woh aur zyada positive ho gayi. Har result ek hi baat kehta hai: ultraviolet light zinc se negative charge bahar nikaalti hai.

2/3Electrons plate chhod dete hain; charge khatam hone par leaves aapas mein gir jaati hain.

Fig. 11.3Ultraviolet light zinc se electrons nikaalti hai: negative plate discharge hoti hai aur neutral plate positive ho jaati hai.

Ek glass case ke andar gold-leaf electroscope ki rod par ek zinc plate rakhi hai, aur top left par ek ultraviolet lamp hai. Step 1: lamp off hai, plate par minus signs hain aur do leaves alag khadi hain, un par bhi minus hai. Step 2: ultraviolet rays plate tak pahunchti hain, chhote electrons usse udte hain jinke arrows par electrons leave likha hai, aur leaves aapas mein gir jaati hain. Step 3: jo plate uncharged thi woh usi light mein hai; ab us par plus signs hain aur plus wali leaves phir alag khadi hain.

Philipp Lenard (1862–1947) ne yeh measurements vacuum ke andar kiye. Unhone ek evacuated glass tube mein do metal plates seal ki aur ek plate, emitter, par ultraviolet light daali, jabki doosri, collector, ko positive rakha. Jab tak light on rahi, theek utni der circuit mein current behta raha. Emitter par free hue electrons vacuum ke across positive collector ki taraf kheenche gaye, yaani metal par padti light ek external circuit mein current chala sakti thi. Unke baad ke measurements ne is photocurrent se teen sawaal puche: yeh collector ke potential par, light ki frequency par, aur uski brightness par kaise respond karta hai?

Ek result sabse alag tha. Ek certain frequency se kam ki light ke liye koi electron nikla hi nahi, plate chahe kitni bhi der roshan rahe. Yeh minimum frequency, threshold frequency (देहली आवृत्ति), emitter ki property hai, aur isi se samajh aata hai ki metals itna alag behave kyun karte hain. Zinc, cadmium ya magnesium ke liye yeh ultraviolet mein hoti hai, isliye visible light kuch nahi karti. Alkali metals (lithium, sodium, potassium, rubidium, caesium) ke liye yeh visible mein hoti hai, isliye ordinary lamp ki light bhi electrons free kar deti hai.

Definition

Photoelectric effect

Kisi surface, aam taur par metal, se electrons ka nikalna jab us par threshold value se zyada frequency ki light padti hai. Jo surfaces yeh dikhati hain unhe photosensitive kehte hain; nikalne wale electrons photoelectrons hain.

11.4 Photoelectric effect ki experimental study

In observations ko laws mein badalne ke liye humein ek aisi tube chahiye jismein har variable set aur measure kiya ja sake. Fig. 11.4 ek design dikhata hai. Glass bulb evacuated hai, taaki electrons raaste mein gas molecules se na takrayein. Andar ek photosensitive emitter C hai, yahan ek curved sheet, aur ek collector A, yahan ek rod. Light ek quartz window se andar aati hai, kyunki ordinary glass ultraviolet light ko absorb kar leta hai.

Jis quantity ko hum badalna chahte hain uska apna control hai. Colour, yaani frequency: lamp ke saamne ka filter badlo, ya khud lamp. Brightness: lamp ko bench par aage-peeche khiskao, kyunki intensity distance ke saath kam hoti hai. Potential: reversing switch (jise aksar commutator kehte hain) wali ek dc supply collector ko emitter ke relative ek chune hue potential $V$ par rakhti hai, positive ya negative. Do meters circuit poora karte hain: $V$ ke liye voltmeter, aur photocurrent ke liye microammeter, jo shayad hi kabhi kuch microamperes se zyada hota hai.

2/3Positive collector electrons ko attract karta hai: photocurrent behta hai.

Fig. 11.4Ek phototube: light emitter se electrons free karti hai, aur collector ka potential tay karta hai ki unmein se kitne pahunchte hain.NCERT Fig. 11.1 · p. 277

Ek gol evacuated glass bulb mein left par emitter C label wali ek curved emitter sheet aur centre ke paas collector A label wali ek seedhi rod hai. Right par ek lamp ki light ek filter aur bulb ki quartz window se guzar kar emitter par padti hai. Neeche emitter aur collector ek microammeter aur adjustable supply se jude hain, aur supply ke across ek voltmeter hai. Step 1: light emitter par padti hai aur electrons bas uski surface chhodte hain; meter zero padhta hai. Step 2: supply collector ko positive banati hai, arrows electrons ko collector tak jaate dikhate hain aur microammeter ki needle ghoomti hai. Step 3: supply reverse ki jaati hai to collector negative ho jaata hai; zyaadatar electrons mud kar emitter par laut jaate hain, sirf sabse tez electron (dashed) collector tak pahunchta hai, aur needle mushkil se hilti hai.

Ab chaar cheezein ek-ek karke badli ja sakti hain: light ki brightness, uski frequency, collector ka potential aur emitter ka material. Agle teen subsections inhe baari-baari lete hain, baaki sab fixed rakh kar.

11.4.1 Photocurrent par light ki intensity ka effect

Collector ko positive rakho, light ka colour fixed rakho aur sirf intensity badlo. Ek simple tareeka hai lamp ko khiskana: intensity distance ke inverse square ki tarah kam hoti hai, isliye distance aadha karne se light chaar guna intense ho jaati hai. Tab microammeter chaar guna padhta hai (Fig. 11.5). Kai distances ke liye dohrao to readings origin se guzarti ek straight line par aati hain: photocurrent intensity ke directly proportional hai.

Collector tak pahunchne wala har electron same charge $e$ le jaata hai, isliye current electrons ko ginta hai: per second $n$ electrons se $I = n e$ milta hai. Intensity ke proportional current isliye humein batata hai ki photoelectrons free hone ka rate light ki intensity ke proportion mein badhta hai.

3/3Saare points origin se guzarti ek straight line par hain: current $\propto$ intensity.

Fig. 11.5Distance aadha karne se intensity chaar guna hoti hai, aur photocurrent bhi chaar guna ho jaata hai.NCERT Fig. 11.2 · p. 278

Left mein ek phototube aur fixed colour ka ek lamp. Right mein photocurrent vs intensity ka graph. Step 1: lamp distance d par hai aur reading intensity 1 aur current 1 par ek point ke roop mein plot hoti hai. Step 2: lamp aadhe distance par aata hai, intensity 4 ho jaati hai aur current 4 par doosra point dikhta hai. Step 3: origin se guzarti ek straight line dono points se jaati hai, jis par straight line likha hai.

11.4.2 Photoelectric current par potential ka effect

Ab light ki frequency aur intensity dono fix karo aur collector ka potential $V$ badlo. Collector positive rakh kar shuru karo aur $V$ ko step by step badhao. Photocurrent badhta hai, kyunki zyada strong attraction un electrons ko bhi samet leta hai jo warna idhar-udhar chale jaate, aur phir woh level off ho jaata hai. Ek certain potential ke baad $V$ aur badhane se kuch nahi badalta: har emit hua electron pehle se collect ho raha hai. Yeh sabse bada current saturation current (संतृप्त धारा) hai.

$V = 0$ par bhi kuch current behta hai, kyunki electrons apni kinetic energy ke saath emitter chhodte hain aur kai bina madad ke collector tak pahunch jaate hain. Unke paas kitni energy hai yeh jaanne ke liye supply reverse karo taaki collector emitter ke respect mein negative ho. Ab field electrons ko emitter ki taraf waapas dhakelta hai, aur sirf kaafi kinetic energy wale electrons hi potential difference chadh kar collector tak pahunch paate hain. Jaise-jaise collector ko aur negative kiya jaata hai, current tezi se girta hai, aur retarding potential ki ek sharply defined value par zero ho jaata hai.

Definition

Stopping potential

Di gayi frequency ki light ke liye collector ke negative (retarding) potential ka woh magnitude $V_{0}$ jis par photocurrent theek zero ho jaata hai. Ise cut-off potential bhi kehte hain.

Ise aise padho. Retarding potential difference $V$ ke against emitter se collector tak jaane wala electron kinetic energy $e V$ kho deta hai. Saare photoelectrons same energy ke saath nahi nikalte, isliye chhota retarding potential sirf sabse dheeme electrons ko rokta hai. Current tab zero hota hai jab retarding potential sabse tez electrons, yaani maximum kinetic energy $K_{\max}$ wale electrons, ko rokne ke liye theek kaafi ho. Isliye

K_{\max} = e V_{0}

(11.1)

jahan

$K_{\max}$: photoelectrons ki sabse badi kinetic energy
$V_{0}$: stopping potential

$0.8 V$ ke stopping potential ka matlab hai ki sabse tez photoelectrons $0.8 e V$ ke saath nikalte hain: electron volt yeh conversion free mein kar deta hai. Ab same frequency ki zyada bright light ke saath poora run dohrao (Fig. 11.6). Saturation current intensity ke proportion mein badhta hai, kyunki har second zyada electrons emit hote hain. Lekin curve bilkul same stopping potential par zero tak pahunchta hai. Di gayi frequency ke liye stopping potential intensity par depend nahi karta: zyada bright light zyada electrons free karti hai, zyada tez electrons nahi. Maximum kinetic energy light ki frequency aur emitter par depend karti hai, light kitni bright hai is par nahi.

3/3Intensity double karne se saturation current double hota hai; cut-off $- 0.8 V$ par hi rehta hai.

Fig. 11.6Zyada bright light saturation current badhati hai lekin stopping potential ko nahi badalti.NCERT Fig. 11.3 · p. 278

Axes photocurrent vs collector potential dikhate hain, lagbhag minus 2 se plus 5 volts tak. Step 1: positive potentials par current badhta hai aur ek value par level off hota hai jis par saturation likha hai. Step 2: curve ko negative potentials tak badhaya jaata hai, jahan current girta hai aur minus 0.8 volt par zero tak pahunchta hai, jise ek dot aur minus 0.8 V label se mark kiya gaya hai. Step 3: double bright light ka ek dashed curve double saturation current par level off hota hai lekin same minus 0.8 volt par zero tak pahunchta hai.

Example

Stopping potential se speed tak

Ek saaf sodium surface par ultraviolet light padti hai. Jab collector emitter se $1.20 V$ neeche hota hai to photocurrent theek ruk jaata hai. Photoelectrons ki maximum kinetic energy eV aur joules mein nikaalo, aur unki sabse badi speed bhi. ( $m_{e} = 9.11 \times 10^{- 31} k g$ )

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

Quick check

3 mein se 1

Fixed frequency ki light ek emitter par padti hai. Uski intensity double kar di jaati hai. Kya hota hai?

Saturation current aur stopping potential dono double ho jaate hainSaturation current double hota hai; stopping potential nahi badaltaSaturation current nahi badalta; stopping potential double hota haiStopping potential aadha ho jaata hai, kyunki electrons energy baant lete hain

Answer dekho

Answer (B) Saturation current double hota hai; stopping potential nahi badalta

Zyada intense light per second zyada electrons free karti hai, isliye saturation current double hota hai. Sabse tez electrons aur tez nahi hote, isliye

V_{0}

(aur

K_{\max} = e V_{0}

) same rehta hai.

V_{0}

ka brightness ke saath badhna wave picture ki ummeed hai, jise experiment galat saabit karta hai.

11.4.3 Stopping potential par incident radiation ki frequency ka effect

Aakhir mein light ki frequency badlo. Barabar ki comparison ke liye har frequency par intensity tab tak adjust karo jab tak saturation current same na ho jaaye. Tab current–potential curves sirf ek jagah alag hote hain: jahan woh axis se milte hain (Fig. 11.7). Frequency jitni zyada, stopping potential utna bada, yaani current rokne ke liye collector ko utna zyada negative karna padta hai. Zyada frequency ki light zyada tez photoelectrons deti hai.

3/3Aur bhi zyada frequency: cut-off $- 2.7 V$ par hai.

Fig. 11.7Zyada frequency (chhoti wavelength) ki light ko current rokne ke liye bada retarding potential chahiye.NCERT Fig. 11.4 · p. 279

Axes photocurrent vs collector potential dikhate hain. Step 1: 400 nanometre light ka ek curve saturation level tak badhta hai aur minus 0.8 volt par zero tak girta hai. Step 2: 300 nanometre light ka ek dashed curve joda jaata hai; uska saturation level same hai lekin woh minus 1.8 volts par zero tak pahunchta hai. Step 3: 250 nanometre light ka ek dotted curve joda jaata hai; saturation level same, zero minus 2.7 volts par. Ek label par same saturation likha hai.

Ab ek emitter ke liye stopping potential vs frequency plot karo (Fig. 11.8). Points ek straight line par aate hain, aur line frequency axis se ek definite frequency $ν_{0}$ par milti hai jahan stopping potential zero hai. $ν_{0}$ se neeche rokne ko kuch hai hi nahi, kyunki koi electron emit nahi hota. Doosre metal ke liye dohrao to ek aur straight line milti hai, pehli ke parallel lekin axis ko ek alag frequency par kaatti hui.

3/4Zinc zyada threshold wali ek parallel line deta hai.

Fig. 11.8Stopping potential frequency ke saath linearly badhta hai, har metal ke liye same slope wali lines par.NCERT Fig. 11.5 · p. 279

Axes volts mein stopping potential, minus 4 se plus 4 tak, ko ten to the fourteen hertz units mein frequency, 0 se 15 tak, ke against dikhate hain. Step 1: potassium ke paanch measured points. Step 2: unse guzarti ek straight line frequency axis se lagbhag 5.6 par milti hai, jo threshold frequency hai aur nu zero se mark hai. Step 3: zinc ki ek dashed line, pehli ke parallel, axis se lagbhag 10.4 par milti hai. Step 4: dono lines ko dotted lines ke roop mein zero frequency tak peeche badhaya jaata hai, jahan woh potential axis ko minus 2.3 aur minus 4.3 volts par kaatti hain, label minus phi zero over e.

Definition

Threshold frequency

Light ki woh minimum frequency $ν_{0}$ jo kisi di gayi surface se photoelectrons nikaal sakti hai. Isse neeche koi emission nahi hota, light chahe kitni bhi intense ho.

Har surface utni hi aasaani se respond nahi karti. Copper ultraviolet light mein photoelectrons deta hai lekin green ya red light mein bilkul nahi, chahe woh kitni bhi bright ho, jabki selenium copper ya zinc dono se zyada aasaani se respond karta hai.

Ek aur fact picture poori karta hai. Jab bhi frequency threshold se upar hoti hai, light on karte hi emission shuru ho jaata hai, koi aisi delay nahi jo kisi experiment ne detect ki ho, chahe light bahut hi dim ho. Measurements ke hisaab se koi bhi delay lagbhag $10^{- 9} s$ se kam hai.

Yeh experiments milkar, kisi di gayi emitter ke liye, photoelectric emission ke chaar laws dete hain:

Threshold frequency se upar photocurrent light ki intensity ke directly proportional hota hai (Fig. 11.5).
Saturation current intensity ke proportional hota hai, lekin stopping potential, aur isliye $K_{\max}$ , intensity par depend nahi karta (Fig. 11.6).
Ek threshold frequency hoti hai jisse neeche koi photoelectron emit nahi hota, light chahe kitni bhi intense ho. Usse upar stopping potential, aur isliye $K_{\max}$ , frequency ke saath linearly badhta hai (Fig. 11.8).
Emission instantaneous hota hai: light pahunchne ke lagbhag $10^{- 9} s$ ke andar shuru ho jaata hai, bahut dim light mein bhi.

InteractivePhotoelectric lab

Ek simulated phototube jiske left mein emitter aur right mein collector hai, aur ek lamp se roshan hai jiska colour chuni gayi wavelength ke hisaab se badalta hai. Controls se emitter metal (caesium, potassium, zinc ya copper), 200 se 700 nanometres tak wavelength, photons per second ke roop mein brightness, aur minus 5 se plus 5 volts tak collector potential chuna jaata hai. Play button electrons ko tube ke across chalata hai. Readouts photon energy, frequency, maximum kinetic energy, stopping potential aur photocurrent dete hain, aur ek status line batati hai ki kya ho raha hai. Ek graph present settings ke liye photocurrent vs collector potential dikhata hai, aur doosra graph record kiye gaye stopping potentials ko frequency ke against jama karta hai aur slope aur usse nikalne wali Planck's constant ki value batata hai.

Fig. 11.9Ek metal chuno, wavelength, brightness aur collector potential set karo, aur Planck's constant naapne ke liye stopping potentials record karo.

Try karke dekho

Photoelectric lab explore karo

Potassium aur $450 n m$ chuno. Collector potential ko aur aur negative karo aur dekho ki current kahan zero hota hai. Ab sirf brightness badlo: kya woh potential khiskta hai?
Wavelength $650 n m$ par set karo aur brightness $100 %$ tak badhao. Phir bhi current kyun nahi hai? Potassium ke liye kaam karne wali sabse lambi wavelength dhoondo.
Ek metal ke liye chaar ya paanch wavelengths par $V_{0}$ record karo. Slope se $h$ ki kya value milti hai? Zinc ke liye dohrao aur dono lines compare karo.

Chalo explore karein

Ek LED jo sirf kuch colours “dekhti” hai

Tumhe chahiye

ek red LED aur ek blue ya green LED
ek digital multimeter
coloured light: plain red, green ya blue dikhati phone screen, ya coloured cellophane se dhaki torch

Pehle guess karo

Tumhe kaunsi pairing se sabse chhoti reading ki ummeed hai?

blue light mein red LEDred light mein blue LEDred light mein red LED

Multimeter ko uski sabse kam dc voltage range par set karo aur uske leads red LED ki do legs par clip karo. Koi battery use nahi hoti: LED hi detector hai.
Dim kamre mein LED ko coloured light source ke paas pakdo, yaani full-screen colour dikhati phone screen ya coloured cellophane ke peeche torch. Red, green aur blue baari-baari try karo aur har ek ki reading note karo.
Blue (ya green) LED ko detector banakar dohrao.
Apni readings table mein record karo.

Table 11.1

	Red screen par (mV)	Green screen par (mV)	Blue screen par (mV)
Red LED
Blue ya green LED

Tumhari entries apne aap save hoti hain.

Red LED teeno colours mein reading deti hai. Blue LED red light mein lagbhag kuch nahi deti, chahe tum use kitna bhi paas pakdo. Har LED ka ek threshold hota hai: bahut kam frequency ki light kuch nahi karti, uski brightness chahe jo ho. Yeh ek semiconductor effect hai, jo tumhe Semiconductor Electronics mein milega, vacuum mein emission nahi, lekin threshold ki wajah same hai: ek photon ko poori energy deni padti hai.

11.5 Photoelectric effect aur light ki wave theory

1900 tak light ki wave theory par koi shak nahi lagta tha. Yeh interference, diffraction aur polarisation explain karti thi, aur Maxwell ki theory ne light ko electromagnetic wave bataya. Is picture mein light beam ki energy har wavefront par smoothly phaili hoti hai aur lagaataar behti hai, pipe se nikalte paani ki tarah, kankadon ki stream ki tarah nahi. Chalo is picture ko chaar laws par test karte hain.

Zyada bright wave ke electric aur magnetic fields zyada strong hote hain, isliye use har surface electron ko zor se hilana chahiye aur zyada energy deni chahiye. Photoelectrons ki maximum kinetic energy intensity ke saath badhni chahiye. Kyunki energy bas jama hoti jaati hai, agar light kaafi bright ho ya kaafi der chamke to koi bhi frequency kaam karni chahiye, yaani koi threshold nahi hona chahiye. Dono predictions experiments se mel nahi khaati (laws 2 aur 3).

Compare

Wave picture vs experiment

Wave picture kya predict karti hai

Zyada bright light zyada tez electrons deti hai
Light kaafi bright ho to koi bhi frequency kaam karti hai
Dim light: emission shuru hone se pehle lamba intezaar
Frequency ka koi khaas role nahi

Experiments kya dikhate hain

$K_{\max}$ intensity par depend nahi karta
$ν_{0}$ se neeche kisi bhi intensity par emission nahi
Lagbhag $10^{- 9} s$ ke andar emission, light kitni bhi dim ho
$K_{\max}$ frequency ke saath linearly badhta hai

Timing bhi utna hi nuksaan karti hai. Wave apni energy poore roshan area par deti hai, jo bahut saare electrons mein bant jaati hai, isliye har electron ka hissa dheere-dheere pahunchta hai. Neeche ka example intezaar ka estimate lagata hai. Dim light ke liye yeh ghanton mein aata hai, jabki observed delay $10^{- 9} s$ hai. Short mein, wave picture photoemission ke sabse simple facts ko bhi explain nahi kar sakti (Fig. 11.10).

Example

Wave humein kitna intezaar karwaati?

$1.0 \times 10^{- 3} W / m^{2}$ intensity ki dim light potassium par padti hai, jiska work function $2.3 e V$ hai. Maan lo, jaisa wave picture allow karti hai, ki ek electron atom ke size ke patch, yaani $0.10 n m$ radius ke circle, par padne wali saari energy sokh leta hai. $2.3 e V$ jama karne ke liye use kitna intezaar karna padega?

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

3/4Ek photon apni saari energy $h ν$ ek electron ko deta hai, jo turant nikal jaata hai.

Fig. 11.10Wave picture mein har electron dheere-dheere energy jama karta hai; photon picture mein ek photon ek electron ko apni saari energy ek saath de deta hai.

Figure mein do rows hain. Upar wali row, wave picture: wavefronts ek metal surface par aate hain jis par paanch cups paanch electrons ko represent karte hain, har ek mein energy ka level kam hai. Step 2: cups teen-chauthai bhar gaye hain, label dim light: hours. Neeche wali row, photon picture. Step 3: ek single wavy packet metal se takrata hai aur ek electron turant nikal jaata hai, label at once. Step 4: doosra packet aata hai aur doosra electron nikalta hai, jo dikhata hai ki zyada photons matlab zyada electrons.

Quick check

3 mein se 1

Kaunsa observation light ki wave picture se consistent hai?

Photocurrent intensity ke proportional haiEk threshold frequency hoti haiEmission bina kisi delay ke shuru hota hai

K_{\max}

intensity par depend nahi karta

Answer dekho

Answer (A) Photocurrent intensity ke proportional hai

Zyada intense wave zyada energy laati hai aur zyada electrons free kar sakti hai, isliye intensity ke saath current ka badhna dono pictures mein fit hota hai. Baaki teen observations theek wahi hain jo wave picture explain nahi kar sakti.

11.6 Einstein ki photoelectric equation: radiation ka energy quantum

1905 mein Albert Einstein ne ek radical raasta suggest kiya. 1900 mein Max Planck ne paaya tha ki woh garam bodies se nikalne wali radiation ko sirf tabhi explain kar sakte hain jab yeh maana jaaye ki garam body aur uski radiation ke beech energy $h ν$ ke steps mein aati-jaati hai, jahan $ν$ frequency hai aur $h$ nature ka ek naya constant hai, jise ab Planck's constant kehte hain. Einstein aur aage gaye. Unhone suggest kiya ki radiation ki energy khud alag-alag lumps, yaani quanta, se bani hai, jinmein se har ek ki energy $h ν$ hai. Light ek smooth flow ki tarah absorb nahi hoti: metal ka electron ya to ek poora quantum leta hai, ya kuch nahi.

DerivationEinstein ki photoelectric equation

Surface ke paas ka ek electron ek quantum absorb karta hai aur is tarah energy $h ν$ paata hai. Metal chhodne ke liye use kam se kam work function $ϕ_{0}$ kharch karna padta hai. Energy conserve hoti hai, isliye jo bachta hai woh metal ke bahar kinetic energy ke roop mein dikhta hai.

Jin electrons ko nikalne ke liye sirf $ϕ_{0}$ chahiye woh sabse kam tightly bound hote hain, aur woh sabse zyada kinetic energy ke saath nikalte hain:

Einstein's photoelectric equation

K_{\max} = h ν - ϕ_{0}

(11.2)

Neeche ke energy levels wale electrons, ya jo surface tak aate hue collisions mein energy kho dete hain, $ϕ_{0}$ se zyada kharch karte hain aur $K_{\max}$ se kam energy ke saath nikalte hain. Isiliye photoelectrons ki energies ek sharp maximum tak phaili hoti hain.

Fig. 11.11 potassium ka energy budget dikhata hai, jiska work function $2.30 e V$ hai. Is picture mein, ek fixed frequency ki light ke liye, intensity bas quanta ginti hai: har second har square metre par kitne aate hain. Light ko zyada bright karne se zyada quanta aate hain; koi ek quantum bada nahi hota.

3/3 $405 n m$ ka photon $3.06 e V$ laata hai: electron $0.76 e V$ tak ke saath nikalta hai.

Fig. 11.11Photon electron ko tabhi free karta hai jab

h ν \geq ϕ_{0}

; jo bachta hai woh electron ki kinetic energy hai.

Neeche ek band potassium ke sabse energetic electrons ko represent karta hai, aur usse 2.30 electronvolts upar ek dashed line par escape level likha hai, saath mein phi zero mark wala double arrow. Step 1: 1.91 electronvolts ke 650 nanometre photon ka vermillion arrow band se upar uthta hai lekin escape level se neeche ruk jaata hai. Step 2: 2.30 electronvolts ke 539 nanometre photon ka green arrow theek escape level tak pahunchta hai. Step 3: 3.06 electronvolts ke 405 nanometre photon ka purple arrow escape level paar karta hai; usse upar wala hissa, 0.76 electronvolt, shaded hai aur us par K max likha hai.

Ab equation ko chaaron laws mein se har ek par check karo.

$K_{\max}$ frequency tay karti hai, intensity nahi. Har emission ek electron dwara absorb kiya gaya ek quantum hai, isliye Eq. (11.2) se $K_{\max}$ , $ν$ aur metal par depend karta hai aur $ν$ ke saath linearly badhta hai. Per second kitne quanta aate hain, jo intensity naapti hai, woh is mein aata hi nahi.
Threshold hota hai. Kinetic energy negative nahi ho sakti, isliye emission ke liye $h ν \geq ϕ_{0}$ chahiye, yaani $ν \geq ν_{0}$ , jahan $ν_{0}$ neeche Eq. (11.3) se milta hai. Bade work function wale metal ko zyada threshold frequency chahiye. $ν_{0}$ se neeche koi single quantum kaafi bada nahi hota, aur quanta apni energy mila nahi sakte, isliye brightness badhane ya intezaar karne se kuch nahi hota.
Current intensity ke proportional hai. Per second zyada quanta matlab zyada electrons jo ek-ek absorb karte hain, isliye per second zyada photoelectrons (jab $ν > ν_{0}$ ).
Koi delay nahi. Quantum absorb karna ek single, instantaneous event hai. Dim light matlab per second aise kam events, dheeme events nahi, isliye pehle electrons turant dikhte hain.

threshold frequency

ν_{0} = \frac{ϕ_{0}}{h}

(11.3)

Eq. (11.2) ko Eq. (11.1) ke $K_{\max} = e V_{0}$ ke saath milane se ek prediction milti hai jise voltmeter se test kiya ja sakta hai, jo $ν \geq ν_{0}$ ke liye valid hai:

V_{0} = \frac{h}{e} ν - \frac{ϕ_{0}}{e}

(11.4)

jahan

$h / e$: slope, har metal ke liye same
$ϕ_{0} / e$: intercept, jo metal par depend karta hai

Yeh ek strong prediction hai. $V_{0}$ vs $ν$ ka graph ek straight line hona chahiye, uska slope har metal ke liye $h / e$ hona chahiye, use frequency axis ko $ν_{0}$ par kaatna chahiye, aur $ν = 0$ tak peeche badhane par use potential axis ko $- ϕ_{0} / e$ par kaatna chahiye. Fig. 11.8 ka aakhri step phir se dekho: theek yahi observe hota hai.

Sabse sakht test ek aise insaan ne kiya jo ummeed kar rahe the ki equation fail hogi. Robert Millikan ne is par das saal, 1906 se 1916, lagaaye, aur ek aisa apparatus banaya jo vacuum ke andar alkali metals ki taaza, saaf surfaces chheel deta tha. Unki hairaani ke liye graphs seedhe aur parallel aaye, bilkul waise jaise Eq. (11.4) maangti hai. Unke slope ko $e$ se multiply karne par $h$ mila; sodium ke liye yeh us constant se match hua jo Planck ne garam body ki radiation se nikaala tha, jo ek bilkul alag phenomenon hai. Aaj ki value hai

h = 6.626 \times 10^{- 34} J s

Same $h$ tak do independent raaste, aur doosre methods se match karte work functions, ne shak ki bahut kam jagah chhodi. Einstein ke quanta accept ho gaye, haalaanki kai physicists is baat se pareshaan rahe ki wave theory ke liye inka kya matlab hai.

Example

Mercury lamp se h naapna

School laboratory mein filters mercury lamp ki do lines chun kar ek phototube ko roshan karte hain. $365 n m$ wavelength ki light $1.40 V$ ka stopping potential deti hai; $546 n m$ ki light $0.27 V$ deti hai. Planck's constant, emitter ka work function aur uski threshold wavelength nikaalo. $c = 3.00 \times 10^{8} m / s$ aur $e = 1.602 \times 10^{- 19} C$ lo.

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

Try karke dekho

Teen LEDs se h ka estimate lagao

Tumhe chahiye

known wavelengths wali red, green aur blue LEDs
ek 4.5 V battery pack aur ek 1 kΩ potentiometer
ek 330 Ω resistor
ek voltmeter ya multimeter
dekhne ke liye ek kaagaz ki tube

Pehle guess karo

Tumhe lagta hai kis LED ko chamakne se pehle sabse zyada voltage chahiye hoga?

redgreenblue

Potentiometer ke do bahar wale ends ko battery pack ke across jodo. Uske sliding contact se ek lead $330 Ω$ resistor se hokar ek LED tak aur phir battery ke negative terminal tak le jao, aur voltmeter ko LED ke across lagao. Ab knob ghumane se LED ka voltage zero se smoothly badhta hai.
Andhere kamre mein voltage ko zero se dheere-dheere badhao. Kaagaz ki tube se LED ko dekho aur woh voltage $V_{on}$ note karo jis par pehli baar halki si chamak dikhti hai.
Baaki do LEDs ke liye dohrao, har ek ki wavelength uske packet ya data sheet se lo.
Har LED ke liye $e V_{on} \approx h c / λ$ , isliye $h \approx e V_{on} λ / c$ estimate karo. Behtar hai ki $V_{on}$ vs $1 / λ$ plot karo: slope lagbhag $h c / e$ hota hai.

Table 11.2

	Wavelength (nm)	Turn-on voltage (V)	h ka estimate (J s)
Red
Green
Blue

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

4.5 e V

energy ke photons ek surface par padte hain jiska work function

2.0 e V

hai. Stopping potential kitna hai, volts mein?

Answer dekho

Answer 2.5 V

K_{\max} = 4.5 - 2.0 = 2.5 e V

, aur

V_{0} = K_{\max} / e = 2.5 V

11.7 Light ki particle nature: photon

$h ν$ ke lumps mein energy dena apne aap light ko particles ki stream nahi bana deta. To kya banayega? Cricket ball energy ke saath momentum bhi le jaati hai aur kisi cheez se takrane par dono de deti hai. Maxwell ki theory pehle se kehti hai ki energy $E$ le jaane wali electromagnetic wave momentum $E / c$ bhi le jaati hai (Part I, Chapter 8). Einstein ne ise ek single quantum par lagaya: har quantum ko momentum $h ν / c$ le jaana chahiye. Wavelength ke terms mein,

E = h ν = \frac{h c}{λ}, p = \frac{E}{c} = \frac{h ν}{c} = \frac{h}{λ}

Definite energy aur definite momentum ke saath har quantum particle ki tarah behave karta hai; iske liye photon naam 1926 mein bana. Pakka saboot 1920s ki shuruaat mein Arthur Compton se aaya, jinhone X-rays ko electrons se scatter kiya. Scattered X-rays ki wavelengths lambi thi, theek utni jitni tab expect hoti jab har photon ek electron se waise takraya ho jaise ek billiard ball doosri se, energy aur momentum baant kar. Einstein ko 1921 ka Nobel Prize mila, Millikan ko 1923 ka (electron ke charge aur apne photoelectric kaam ke liye), aur Compton ko 1927 ka.

Photon picture kya kehti hai, property by property, yeh rahi.

Energy aur momentum. Frequency $ν$ ka photon $E = h ν$ aur $p = h ν / c = h / λ$ le jaata hai, aur vacuum mein hamesha $c$ se chalta hai.
Identical photons. Ek single frequency ki light identical photons se bani hoti hai. Brightness batati hai ki per second har square metre se kitne guzarte hain, yeh nahi ki har ek kitna bada hai.
Koi charge nahi. Photons electrically neutral hote hain, isliye na electric aur na magnetic fields unke paths ko mod sakte hain.
Collisions. Jab photon kisi electron ya doosre particle se milta hai, to energy aur momentum dono conserve hote hain, lekin photons ki ginti nahi: photons absorb ho sakte hain, aur naye ban sakte hain.
Kahan lagu hota hai. Yeh particle properties tab dikhti hain jab light matter se emit, absorb ya scatter hoti hai; jab woh travel karti hai, tab interference ke liye wave description hi chahiye.

2/3Bright red light: bahut zyada photons, lekin har ek wahi red photon.

Fig. 11.12Brightness tay karti hai ki har second kitne photons aate hain; frequency tay karti hai ki har ek kitni energy laata hai.

Left par ek lamp right par ek detector ki taraf wave packets bhejta hai. Step 1: dim red light, sirf teen packets, label few photons. Step 2: bright red light, bilkul same size aur wavelength ke nau packets, label many photons. Step 3: dim blue light, saaf taur par chhoti wavelength wale teen packets, label few, bigger photons. Neeche ek formula hai: E equals h nu, p equals h over lambda.

Example 11.1NCERT Example 11.1 · p. 284

Water purifier se photons

Ghar ke water purifier ka ultraviolet lamp $254 n m$ wavelength par $2.5 W$ radiation deta hai. (a) Ek photon ki energy joules aur electron volts mein kitni hai? (b) Lamp har second kitne photons emit karta hai? (c) Har photon ka momentum kitna hai?

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

Example 11.2NCERT Example 11.2 · p. 284

Kya red light kaam karegi?

Potassium ka work function $2.30 e V$ hai. (a) Iski threshold frequency aur threshold wavelength nikaalo. (b) $405 n m$ ke violet laser ki light is par padti hai. Photoelectrons ki maximum kinetic energy, stopping potential aur electrons ki sabse badi speed nikaalo. (c) Kya $650 n m$ ki light koi electron nikaalegi? ( $h = 6.626 \times 10^{- 34} J s$ , $c = 3.00 \times 10^{8} m / s$ , $m_{e} = 9.11 \times 10^{- 31} k g$ )

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

Quick check

4 mein se 1

620 n m

wavelength ki orange light ke photon ki energy electron volts mein kitni hai? (

h c = 1240 e V n m

)

Answer dekho

Answer 2 eV

E = h c / λ = 1240 / 620 = 2.00 e V

11.8 Matter ki wave nature

Is chapter aur Wave Optics par peeche mud kar dekho, to lagta hai light experiment ke hisaab se apna character badal leti hai. Fringes aur diffraction patterns sirf waves se ban sakte hain; photoelectric effect aur Compton scattering, jinmein light energy aur momentum deti hai, sirf photons se explain hote hain. Aksar ek hi device ko dono descriptions ek saath chahiye. Phone camera mein lens image banata hai, aur woh image kitni sharp ho sakti hai yeh diffraction, ek wave effect, se limit hota hai. Lekin sensor ka har pixel photoelectric idea par kaam karta hai: aane wale photons silicon ke andar electrons free karte hain, ek photon se zyada se zyada ek electron, aur camera unhe ginta hai (Fig. 11.13).

2/2Har pixel mein aane wala photon ek electron free karta hai; camera count padhta hai.

Fig. 11.13Camera dono pictures use karta hai: lens wave ki tarah image banata hai, aur sensor photons ko unke free kiye electrons se ginta hai.

Left mein ek camera body jiske aage lens aur peeche sensor chip hai. Step 1: left se teen rays aati hain aur lens unhe mod deta hai taaki woh sensor par ek point par milein. Step 2: sensor ka ek magnified circle nau pixels ka grid dikhata hai; do wave packets, yaani photons, do pixels par aate hain, aur un dono pixels mein ab ek free hua electron hai, jo dot se dikhaya gaya hai, ginne ke liye tayyar.

Louis de Broglie ne 1924 mein argument ko ulta ghuma diya. Light, jise lambe samay se wave kaha jaata tha, particle-sized parcels mein momentum le jaati nikli; shayad electrons, jinhe lambe samay se particles kaha jaata tha, ek wave le jaate hon. Unki wajah symmetry mein vishwas tha. Matter aur radiation universe ke do bade ingredients hain, aur yeh unlikely lagta tha ki sirf ek hi double life jiye. Unhone propose kiya ki momentum $p$ wale har particle ke saath ek wave hoti hai jiski wavelength hai

de Broglie relation

λ = \frac{h}{p} = \frac{h}{m v}

(11.5)

jahan

$m$: particle ka mass
$v$: uski speed

Equation (11.5) de Broglie relation hai, aur $λ$ particle ki de Broglie wavelength (दे ब्रॉग्ली तरंगदैर्ध्य) hai, yaani uski matter wave ki wavelength. Dono sides ko alag-alag padho to duality saaf dikhti hai: $λ$ aisi cheez hai jo sirf wave ki ho sakti hai, $p$ aisi cheez jise hum ek chalte lump ki maante hain, aur chhota sa constant $h$ ek ko doosre mein convert karta hai.

Electrons aur atoms ke liye 1924 mein yeh relation ek bold guess tha jo experiment ka intezaar kar raha tha. Light ke liye ise turant check kiya ja sakta hai. Frequency $ν$ ke photon ka momentum hai

p = \frac{h ν}{c}

(11.6)

isliye de Broglie ka recipe use wavelength deta hai

λ = \frac{h}{p} = \frac{c}{ν}

(11.7)

jo light ki jaani-pehchaani wavelength hi hai. Naye rule mein purana rule shaamil hai.

Hum chalti cheezon ki waves kabhi notice kyun nahi karte? Kyunki $h$ itna chhota hai ki wavelength bahut hi chhoti hoti hai jab tak mass aur speed bhi bahut chhote na hon. $9 m / s$ se girti $4 m g$ mass ki baarish ki boond ka $λ = h / m v \approx 1.8 \times 10^{- 29} m$ hai, jo proton se lagbhag $10^{14}$ guna chhota hai. Koi slit ya grating ise kabhi nahi dikha sakti. Electron ki kahani alag hai: uska mass itna chhota hai ki uski de Broglie wavelength crystal mein atoms ki spacing jitni hoti hai, aur tab wave behaviour real aur measurable hota hai (Fig. 11.14). Yeh bhi note karo ki $λ$ sirf momentum par depend karti hai, particle ke charge ya type par nahi.

4/4Yeh atoms ka size hai, isliye crystals electrons ko diffract kar sakte hain.

Fig. 11.14Roz ki cheezon ke liye matter waves itni chhoti hain ki notice hi nahi hoti; electrons ke liye woh lagbhag atoms ke size ki hain.

Wavelength ka ek horizontal logarithmic axis ten to the minus 38 se ten to the minus 4 metres tak jaata hai, har chhe powers of ten par tick labels ke saath. Step 1: ek sprinter ka marker, lagbhag ten to the minus 36 metres par, aur girti baarish ki boond ka, lagbhag ten to the minus 29 metres par. Step 2: dheere tairta ek pollen grain, lagbhag ten to the minus 19 metres par. Step 3: 150 volts se accelerate hua electron, ten to the minus 10 metres par. Step 4: atoms ke size ka ek shaded band electron ke marker ko cover karta hai, aur visible light ka ek patla band aur right mein hai.

DerivationPotential difference se accelerate hue electron ki wavelength

Rest se start hokar potential difference $V$ se accelerate hua electron kinetic energy $K = e V$ paata hai. Kyunki $K = p^{2} / 2 m$ , uska momentum $p = \sqrt{2 m e V}$ hai, aur Eq. (11.5) deti hai

λ = \frac{h}{\sqrt{2 m e V}}

Electron ke liye $h$ , $m$ aur $e$ daalne par, $V$ volts mein lekar,

λ = \frac{1.227}{\sqrt{V}} n m

$V = 150 V$ ke liye isse $λ = 0.100 n m$ milta hai, jo crystal mein paas-paas ke atoms ke beech ki doori jitna hai. (Formula relativity ko ignore karta hai, jo kuch hazaar volts tak theek hai.)

Example 11.3NCERT Example 11.3 · p. 286

Teen de Broglie wavelengths

In ki de Broglie wavelength nikaalo: (a) $1.2 \times 10^{7} m / s$ se chalta electron, (b) $480 m / s$ se chalta oxygen molecule ( $m = 5.3 \times 10^{- 26} k g$ ), jo hawa mein typical speed hai, aur (c) $10 m / s$ se daudta $70 k g$ ka sprinter. ( $h = 6.626 \times 10^{- 34} J s$ , $m_{e} = 9.11 \times 10^{- 31} k g$ )

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

Example

Diffraction experiment ke liye electrons

Electrons ko ek crystal se diffract karna hai jiske atomic planes $0.20 n m$ door hain, isliye unki de Broglie wavelength bhi lagbhag $0.20 n m$ honi chahiye. Unhe rest se kitne potential difference se accelerate karna chahiye?

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

Recap

Summary aur key points

Cathode rays electrons ki streams hain. Crossed fields se unka specific charge mila, $e / m = 1.76 \times 10^{11} C / k g$ , jo har source ke liye same hai; Millikan ki oil drops se $e = 1.602 \times 10^{- 19} C$ mila (Section 11.1).
Work function $ϕ_{0}$ woh sabse kam energy hai jo metal surface se electron ko free karti hai. Heating, strong field ya light ise de sakti hai; $1 e V = 1.602 \times 10^{- 19} J$ (Section 11.2).
Photoelectric effect mein kaafi zyada frequency ki light surface se electrons nikaalti hai. Kuch metals (zinc, copper) ko ultraviolet light chahiye; alkali metals visible light par respond karte hain (Section 11.3).
Photocurrent light ki intensity ke proportional hai, kyunki per second emit hone wale electrons ki sankhya bhi proportional hai (Section 11.4.1).
Collector ka potential badhane se current ek saturation value tak badhta hai. Retarding potential ise stopping potential $V_{0}$ par zero tak le aata hai, jahan $e V_{0} = K_{\max}$ ; $V_{0}$ intensity par depend nahi karta (Section 11.4.2).
$V_{0}$ frequency ke saath linearly badhta hai; threshold frequency $ν_{0}$ se neeche kisi bhi intensity par emission nahi hota, aur usse upar emission instantaneous hota hai (Section 11.4.3).
Light ki wave picture predict karti hai ki $K_{\max}$ intensity ke saath badhega, koi threshold nahi hoga, aur dim light delay karegi. Teeno galat hain (Section 11.5).
Einstein: light energy quanta $h ν$ mein aati hai, aur ek electron ek quantum absorb karta hai, isliye $K_{\max} = h ν - ϕ_{0}$ aur $ν_{0} = ϕ_{0} / h$ (Section 11.6).
Isliye $V_{0} = (h / e) ν - ϕ_{0} / e$ : har metal ke liye slope $h / e$ wali straight line. Millikan ke measurements ne ise confirm kiya aur $h = 6.626 \times 10^{- 34} J s$ diya.
Photon ki energy $E = h ν$ aur momentum $p = h ν / c = h / λ$ hota hai, us par koi charge nahi hota aur woh $c$ se chalta hai. Intensity photons ki sankhya tay karti hai, unki energy nahi (Section 11.7).
Light ki koi bhi picture akele complete nahi hai: fringes ko waves chahiye, photoemission ko photons, aur camera jaisa device dono use karta hai (Section 11.8).
De Broglie: momentum $p$ wale particle ki wavelength $λ = h / p = h / m v$ hoti hai, jo uske charge se independent hai. Electrons ke liye yeh measurable hai (lagbhag atoms ki spacing jitni) aur roz ki cheezon ke liye bahut hi chhoti (Section 11.8).

Table 11.3Is chapter mein mili quantities

Quantity	Symbol	SI unit	Dimensions	Kahan aati hai
Planck's constant	$h$	$J s$	$[M L^{2} T^{- 1}]$	$E = h ν$ , $p = h / λ$
Work function	$ϕ_{0}$	J (aksar eV)	$[M L^{2} T^{- 2}]$	$K_{\max} = h ν - ϕ_{0}$
Threshold frequency	$ν_{0}$	Hz	$[T^{- 1}]$	$ν_{0} = ϕ_{0} / h$
Stopping potential	$V_{0}$	V	$[M L^{2} T^{- 3} A^{- 1}]$	$e V_{0} = K_{\max}$
Photon momentum	$p$	$k g m / s$	$[M L T^{- 1}]$	$p = h ν / c$
de Broglie wavelength	$λ$	m	$[L]$	$λ = h / p$

Flash cards · 9

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Practice · 15 questions

Practice questions answers ke saath (NCERT “Exercises”)

1
NumericalEasyNCERT Exercises 11.1
Ek dental X-ray machine electrons ko tungsten target se takrane se pehle $65 k V$ se accelerate karti hai. Jo electron ek hi collision mein ruk jaata hai woh apni saari kinetic energy ek X-ray photon ko de sakta hai.
1. Bane X-rays mein sabse zyada frequency kitni hai?
  Hz
  Solution dikhao
  $h ν_{\max} = e V \Rightarrow ν_{\max} = \frac{e V}{h} = \frac{(1.602 \times 10^{- 19} C) (6.5 \times 10^{4} V)}{6.626 \times 10^{- 34} J s} = 1.57 \times 10^{19} H z$
2. Sabse chhoti wavelength nanometres mein kitni hai?
  nm
  Solution dikhao
  $λ_{\min} = \frac{c}{ν_{\max}} = \frac{3.00 \times 10^{8} m / s}{1.57 \times 10^{19} H z} = 1.91 \times 10^{- 11} m = 0.0191 n m$
  Isi tarah, $λ_{\min} = 1240 e V n m / 65 000 e V = 0.0191 n m$ .
2
NumericalMediumNCERT Exercises 11.2
Lithium ka work function $2.90 e V$ hai. $350 n m$ wavelength ki ultraviolet light lithium surface par padti hai.
1. Photoelectrons ki maximum kinetic energy eV mein nikaalo.
  eV
  Solution dikhao
  $h ν = \frac{h c}{λ} = \frac{1240 e V n m}{350 n m} = 3.55 e V$
  $K_{\max} = h ν - ϕ_{0} = 3.55 e V - 2.90 e V = 0.65 e V$
2. Stopping potential kitna hai?
  V
  Solution dikhao
  $e V_{0} = K_{\max}$ , isliye eV mein $K_{\max}$ ho to volts mein stopping potential ka number same hota hai: $V_{0} = 0.65 V$ .
3. Photoelectrons ki sabse badi speed kitni hai?
  m/s
  Solution dikhao
  $v_{\max} = \sqrt{\frac{2 K_{\max}}{m_{e}}} = \sqrt{\frac{2 \times 0.645 \times 1.602 \times 10^{- 19} J}{9.11 \times 10^{- 31} k g}} = 4.8 \times 10^{5} m / s$
3
NumericalEasyNCERT Exercises 11.3
Ek photoelectric experiment mein current tab zero ho jaata hai jab collector emitter ke respect mein $0.85 V$ negative hota hai.
1. Photoelectrons ki maximum kinetic energy joules mein nikaalo.
  J
  Solution dikhao
  $K_{\max} = e V_{0} = (1.602 \times 10^{- 19} C) (0.85 V) = 1.36 \times 10^{- 19} J = 0.85 e V$
2. Sabse tez photoelectrons kitni speed se chal rahe hain?
  m/s
  Solution dikhao
  $v_{\max} = \sqrt{\frac{2 \times 1.36 \times 10^{- 19} J}{9.11 \times 10^{- 31} k g}} = 5.5 \times 10^{5} m / s$
4
NumericalMediumNCERT Exercises 11.4
Optical mouse ka ek laser diode $780 n m$ wavelength ki $4.5 m W$ infrared light emit karta hai.
1. Har photon ki energy (joules aur eV mein) aur momentum nikaalo.
  Solution dikhao
  Answer $E = 2.55 \times 10^{- 19} J = 1.59 e V$ ; $p = 8.49 \times 10^{- 28} k g m / s$ .
  $E = \frac{h c}{λ} = \frac{(6.626 \times 10^{- 34}) (3.00 \times 10^{8})}{780 \times 10^{- 9}} J = 2.55 \times 10^{- 19} J = 1.59 e V$
  $p = \frac{h}{λ} = \frac{6.626 \times 10^{- 34} J s}{780 \times 10^{- 9} m} = 8.49 \times 10^{- 28} k g m / s$
2. Average mein har second diode se kitne photons nikalte hain?
  photons per second
  Solution dikhao
  $N = \frac{P}{E} = \frac{4.5 \times 10^{- 3} W}{2.55 \times 10^{- 19} J} = 1.8 \times 10^{16} per second$
3. Ek neutron ( $m = 1.675 \times 10^{- 27} k g$ ) ko in photons mein se ek ke barabar momentum ke liye kitni tez chalna padega?
  m/s
  Solution dikhao
  $v = \frac{p}{m} = \frac{8.49 \times 10^{- 28} k g m / s}{1.675 \times 10^{- 27} k g} = 0.51 m / s$
  Paidal chalne se bhi dheere: ek single photon ka momentum bahut chhota hota hai.
5
NumericalMediumNCERT Exercises 11.5
Ek student ek emitter ke liye stopping potentials naapta hai: $5.5 \times 10^{14} H z$ par $0.17 V$ , $7.0 \times 10^{14} H z$ par $0.79 V$ , $8.5 \times 10^{14} H z$ par $1.41 V$ aur $10.0 \times 10^{14} H z$ par $2.03 V$ .
1. $V_{0}$ vs $ν$ graph ka slope $10^{- 15} V s$ ki units mein kitna hai?
  × 10⁻¹⁵ V s
  Solution dikhao
  Har $1.5 \times 10^{14} H z$ par points $0.62 V$ badhte hain, isliye woh ek straight line par hain. End points use karke:
  $slope = \frac{2.03 V - 0.17 V}{10.0 \times 10^{14} H z - 5.5 \times 10^{14} H z} = 4.13 \times 10^{- 15} V s$
2. Isse Planck's constant ki kya value milti hai?
  J s
  Solution dikhao
  $h = e \times slope = (1.602 \times 10^{- 19} C) (4.13 \times 10^{- 15} V s) = 6.62 \times 10^{- 34} J s$
3. Emitter ki threshold frequency aur work function nikaalo.
  Solution dikhao
  Answer $ν_{0} \approx 5.1 \times 10^{14} H z$ ; $ϕ_{0} \approx 2.1 e V$ .
  Line pehle point se $0.17 V$ neeche $V_{0} = 0$ tak girti hai:
  $ν_{0} = 5.5 \times 10^{14} H z - \frac{0.17 V}{4.13 \times 10^{- 15} V s} = 5.1 \times 10^{14} H z$
  $ϕ_{0} = h ν_{0} = \frac{(6.62 \times 10^{- 34}) (5.09 \times 10^{14})}{1.602 \times 10^{- 19}} e V = 2.1 e V$
NumericalEasyNCERT Exercises 11.6
Ek photosensitive surface ki threshold wavelength $620 n m$ hai. Us par $410 n m$ wavelength ki violet light padti hai. Kitna stopping potential chahiye?
V
Solution dikhao
$ϕ_{0} = \frac{h c}{λ_{0}} = \frac{1240 e V n m}{620 n m} = 2.00 e V, h ν = \frac{1240 e V n m}{410 n m} = 3.02 e V$
$e V_{0} = h ν - ϕ_{0} = 3.02 e V - 2.00 e V = 1.02 e V \Rightarrow V_{0} \approx 1.0 V$
Short answerEasyNCERT Exercises 11.7
Tungsten ka work function $4.5 e V$ hai. Ek lamp $280 n m$ wavelength ki ultraviolet light deta hai. Kya yeh tungsten surface se photoelectrons nikaalega? Explain karo.
Solution dikhao
AnswerNahi: har photon mein sirf $4.43 e V$ hai, jo $4.5 e V$ work function se thoda kam hai.
$h ν = \frac{h c}{λ} = \frac{1240 e V n m}{280 n m} = 4.43 e V < 4.5 e V$
Kisi single photon mein kaafi energy nahi hai, isliye koi electron emit nahi hota, lamp chahe kitna bhi intense ho. Tungsten ki threshold wavelength $1240 / 4.5 = 276 n m$ hai; sirf isse chhoti wavelengths kaam karti hain.
NumericalMediumNCERT Exercises 11.8
Ek ultraviolet LED se $365 n m$ wavelength ki light ek metal par padti hai, aur sabse tez photoelectrons $4.4 \times 10^{5} m / s$ se nikalte hain. Metal ki threshold frequency nikaalo.
Hz
Solution dikhao
$ν = \frac{c}{λ} = \frac{3.00 \times 10^{8} m / s}{365 \times 10^{- 9} m} = 8.22 \times 10^{14} H z$
$K_{\max} = \frac{1}{2} m_{e} v^{2} = \frac{1}{2} (9.11 \times 10^{- 31} k g) (4.4 \times 10^{5} m / s)^{2} = 8.82 \times 10^{- 20} J$
Eq. (11.2) se, $h ν_{0} = ϕ_{0} = h ν - K_{\max}$ , isliye
$ν_{0} = ν - \frac{K_{\max}}{h} = 8.22 \times 10^{14} H z - \frac{8.82 \times 10^{- 20} J}{6.626 \times 10^{- 34} J s} = 6.9 \times 10^{14} H z$
NumericalEasyNCERT Exercises 11.9
$445 n m$ ka blue laser ek phototube ke emitter par chamakta hai, aur stopping potential $0.62 V$ hai. Emitter ka work function kitna hai?
eV
Solution dikhao
$h ν = \frac{1240 e V n m}{445 n m} = 2.79 e V$
$ϕ_{0} = h ν - e V_{0} = 2.79 e V - 0.62 e V = 2.17 e V$
Yeh caesium ke work function ke kareeb hai.
10
NumericalEasyNCERT Exercises 11.10
In mein se har ek ki de Broglie wavelength nikaalo.
1. $20 m / s$ se chalti $1500 k g$ ki car.
  m
  Solution dikhao
  $λ = \frac{h}{m v} = \frac{6.626 \times 10^{- 34} J s}{(1500 k g) (20 m / s)} = 2.2 \times 10^{- 38} m$
2. $0.50 m / s$ se udta $2.5 m g$ mass ka mosquito.
  m
  Solution dikhao
  $λ = \frac{6.626 \times 10^{- 34} J s}{(2.5 \times 10^{- 6} k g) (0.50 m / s)} = 5.3 \times 10^{- 28} m$
3. $1.0 m m / s$ se tairta $1.0 \times 10^{- 15} k g$ mass ka smoke particle.
  m
  Solution dikhao
  $λ = \frac{6.626 \times 10^{- 34} J s}{(1.0 \times 10^{- 15} k g) (1.0 \times 10^{- 3} m / s)} = 6.6 \times 10^{- 16} m$
  Teeno mein sabse badi yeh wavelength bhi atomic nucleus se chhoti hai, observe karne ke liye bahut hi chhoti.
ProofMediumNCERT Exercises 11.11
Photon ke liye $E = h ν$ aur is fact se shuru karo ki light momentum $p = E / c$ le jaati hai, aur dikhao ki photon ke liye $h / p$ light ki wavelength ke barabar hai. Phir is result se $600 n m$ wavelength ke photon ka momentum nikaalo.
Solution dikhao
Answer $h / p = λ$ ; $600 n m$ par $p = 1.10 \times 10^{- 27} k g m / s$ .
$p = \frac{E}{c} = \frac{h ν}{c} \Rightarrow \frac{h}{p} = \frac{c}{ν} = λ$
To photon ki de Broglie wavelength us radiation ki wavelength hai jiska woh hissa hai.
$p = \frac{h}{λ} = \frac{6.626 \times 10^{- 34} J s}{600 \times 10^{- 9} m} = 1.10 \times 10^{- 27} k g m / s$
Quick answerEasy
Threshold se upar, ek single frequency ki light ek metal par padti hai. Kaunsa change per second emit hone wale photoelectrons ki sankhya badhayega?
Light ki intensity badhanaCollector ko aur negative karnaBade work function wala metal use karnaLamp ko aur door le jaana
Answer dekho
Answer (A) Light ki intensity badhana
Zyada intense light per second zyada photons laati hai, isliye zyada electrons free hote hain. Zyada negative collector emission nahi badalta (woh sirf electrons ko waapas modta hai), aur baaki do emit hone wale electrons ki sankhya kam karte hain.
Quick answerMedium
$1.0 e V$ energy ke photons ki wavelength nm hoti hai (nearest 10 nm tak).
Answer dekho
Answer 1240
$λ = h c / E = 1240 e V n m / 1.0 e V = 1240 n m$ , jo infrared mein hai.
Quick answerMedium
Ek electron aur ek proton ki kinetic energy same hai. Electron ki de Broglie wavelength lagbhag kitne guna lambi hai?
guna
Answer dekho
Answer 42.85 guna
$λ_{e} / λ_{p} = \sqrt{m_{p} / m_{e}} = \sqrt{1836} \approx 43$ .
15
Quick answerEasy
Har observation ko us idea se match karo jo use explain karta hai.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
Threshold frequency : ek photon ko kam se kam $ϕ_{0}$ dena hoga
Koi time lag nahi : photon ek single event mein absorb hota hai
Current intensity ke proportional : zyada photons zyada electrons free karte hain
Crystals se electron diffraction : electrons ki wavelength $h / p$ hoti hai
Har photoelectric law ek-photon, ek-electron absorption se nikalta hai; electron diffraction de Broglie ki matter waves ka saboot hai.

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 11.1	Practice Q1
NCERT Exercises 11.2	Practice Q2
NCERT Exercises 11.3	Practice Q3
NCERT Exercises 11.4	Practice Q4
NCERT Exercises 11.5	Practice Q5
NCERT Exercises 11.6	Practice Q6
NCERT Exercises 11.7	Practice Q7
NCERT Exercises 11.8	Practice Q8
NCERT Exercises 11.9	Practice Q9
NCERT Exercises 11.10	Practice Q10
NCERT Exercises 11.11	Practice Q11

Quick check: MCQs

11.1 Introduction3 questions
11.2 Electron emission3 questions
11.3.2 Hallwachs aur Lenard ke observations3 questions
11.4.2 Photoelectric current par potential ka effect3 questions
11.4.3 Stopping potential par incident radiation ki frequency ka effect3 questions
11.5 Photoelectric effect aur light ki wave theory3 questions
11.6 Einstein ki photoelectric equation: radiation ka energy quantum3 questions
11.7 Light ki particle nature: photon4 questions
11.8 Matter ki wave nature4 questions

Glossary · 15

Key terms (English + Hindi)

cathode raysकैथोड किरणेंkaithod kiranen: Bahut kam pressure par gas-discharge tube ke cathode se nikalne wali electrons ki streams.Chapter mein pehli baar yahan aaya →

de Broglie wavelengthदे ब्रॉग्ली तरंगदैर्ध्यde broglie tarangdairghya: Momentum $p$ wale chalte particle se judi wave ki wavelength $λ = h / p$ .Chapter mein pehli baar yahan aaya →
dual natureद्वैत प्रकृतिdvait prakriti: Yeh fact ki radiation aur matter dono experiment ke hisaab se wave jaisa aur particle jaisa behaviour dikhate hain.Chapter mein pehli baar yahan aaya →

electron volt (eV)इलेक्ट्रॉन वोल्टelectron volt: Woh energy jo electron $1 V$ se accelerate hone par paata hai: $1 e V = 1.602 \times 10^{- 19} J$ .Chapter mein pehli baar yahan aaya →

field emissionक्षेत्र उत्सर्जनkshetra utsarjan: $10^{8} V / m$ order ke bahut strong electric field se metal se electrons ka nikalna.Chapter mein pehli baar yahan aaya →

photoelectric effectप्रकाश-विद्युत प्रभावprakash-vidyut prabhav: Kaafi zyada frequency ki light padne par surface se electrons ka emission.
photoelectronप्रकाशिक इलेक्ट्रॉनprakashik electron: Light dwara surface se free kiya gaya electron.Chapter mein pehli baar yahan aaya →
photonफ़ोटॉनphoton: Electromagnetic radiation ka ek quantum, jiski energy $h ν$ aur momentum $h / λ$ hai, aur jo $c$ se chalta hai.Chapter mein pehli baar yahan aaya →
Planck's constantप्लांक नियतांकplank niyatank: Constant $h = 6.626 \times 10^{- 34} J s$ jo quantum ki energy ko uski frequency se jodta hai.Chapter mein pehli baar yahan aaya →

saturation currentसंतृप्त धाराsantript dhara: Di gayi light setting par sabse bada photocurrent, jo tab milta hai jab har emit hua electron collector tak pahunchta hai.Chapter mein pehli baar yahan aaya →
specific chargeविशिष्ट आवेशvishisht avesh: Particle ka charge divided by uska mass; electron ke liye $e / m = 1.76 \times 10^{11} C / k g$ .Chapter mein pehli baar yahan aaya →
stopping potentialनिरोधी विभवnirodhi vibhav: Collector ka woh retarding potential $V_{0}$ jis par photocurrent theek zero ho jaata hai; $e V_{0} = K_{\max}$ .

thermionic emissionतापायनिक उत्सर्जनtapayanik utsarjan: Metal ko itna garam karne par electrons ka nikalna ki kuch free electrons ko work function se zyada energy mil jaaye.Chapter mein pehli baar yahan aaya →
threshold frequencyदेहली आवृत्तिdehali avritti: Sabse kam frequency $ν_{0} = ϕ_{0} / h$ jo kisi surface se electrons nikaal sakti hai, light chahe kitni bhi intense ho.Chapter mein pehli baar yahan aaya →

work functionकार्य फलनkarya phalan: Metal ki surface se electron ko free karne ke liye zaroori sabse kam energy, $ϕ_{0}$ ; yeh metal aur uski surface par depend karta hai.Chapter mein pehli baar yahan aaya →

11.1 Introduction

11.2 Electron emission

11.3 Photoelectric effect

11.3.1 Hertz ke observations

11.3.2 Hallwachs aur Lenard ke observations

11.4 Photoelectric effect ki experimental study

11.4.1 Photocurrent par light ki intensity ka effect

11.4.2 Photoelectric current par potential ka effect

11.4.3 Stopping potential par incident radiation ki frequency ka effect

11.5 Photoelectric effect aur light ki wave theory

11.6 Einstein ki photoelectric equation: radiation ka energy quantum

11.7 Light ki particle nature: photon

11.8 Matter ki wave nature

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