Class 12 Physics · Chapter 11
Short answer:
Is chapter me NCERT ke exercise se 18 numerical-heavy questions cover kiye gaye hain — electron emission, photoelectric effect ke experimental laws, Einstein's photoelectric equation (KEmax=hν−φ0), stopping potential, photon energy/momentum, aur de Broglie wavelength (matter waves) ke numericals. Har calculation me full working diya hai — given data se substitution tak, correct SI aur eV units ke saath.
Photoelectric effect ne prove kiya ki light sirf wave nahi, particle (photon) ki tarah bhi behave karti hai — kyunki ek photon ek hi electron ko instantly energy deta hai, wave theory isse explain nahi kar pati. Ye chapter light ki dual nature (wave + particle) aur matter ki dual nature (particle + wave, de Broglie hypothesis) dono establish karta hai. Board exam me ye chapter numerical-heavy hai — Einstein's equation, stopping potential aur de Broglie wavelength ke sawaal har saal aate hain, isliye formula aur unit-conversion pe pakad zaroori hai.
Chapter 11 Summary — 5 Minute Revision
1. Electron Emission
Metal ke andar free electrons hote hain jo metal surface ke potential barrier ki wajah se bahar nahi nikal pate. Inhe bahar nikalne ke liye minimum energy chahiye, jise work function (φ0) kehte hain (units: eV ya J).
- Thermionic emission — metal ko garam karke (jaise CRT filament)
- Field emission — strong electric field lagakar
- Photoelectric emission — suitable frequency ki light daalkar (ye chapter isi pe focus karta hai)
2. Photoelectric Effect — Experimental Observations (Laws)
- Photoelectric current intensity ke directly proportional hai (frequency fixed rakho)
- Stopping potential (V0) frequency ke saath badhta hai, intensity se independent hai
- Har metal ke liye ek threshold frequency (ν0) hoti hai — usse kam frequency pe emission bilkul nahi hota, chahe intensity kitni bhi zyada ho
- Emission instantaneous hai — no time lag, jaisi hi threshold se zyada frequency ki light padti hai turant electron nikalte hain
Ye saari observations wave theory se explain nahi hoti (wave theory me energy intensity pe depend karni chahiye, frequency pe nahi) — isliye Einstein ne photon concept diya.
3. Einstein's Photoelectric Equation
Light photons ki stream hai, har photon ki energy E=hν. Ek photon ek hi electron ko poori energy deta hai:
KEmax = hν − φ0 = hν − hν0
Isse saari observations explain hoti hain: agar hν < φ0, koi electron emit nahi hoga (chahe intensity jitni bhi ho) — yehi threshold frequency ka origin hai.
4. Stopping Potential
Fastest photoelectrons ko rokne ke liye jo retarding potential lagana padta hai:
eV0 = KEmax
5. Particle Nature of Light — Photon
| Property | Formula |
|---|---|
| Energy | E = hν = hc/λ |
| Momentum | p = E/c = h/λ |
| Rest mass | zero |
↔ Table ko side me swipe karein
6. Wave Nature of Matter — de Broglie Relation
Jaise light kabhi wave kabhi particle behave karti hai, waise hi matter (electron, proton, etc.) bhi kabhi particle kabhi wave behave karta hai. Kisi bhi moving particle ki associated wavelength:
λ = h/p = h/(mv)
Yaad rakho: agar kinetic energy di ho na ki velocity, to pehle momentum nikalo — p = √(2mKE) — phir λ = h/p use karo.
Poore Class 12 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. Sodium ka work function 2.14 eV hai. Iski threshold frequency kya hogi? Sodium se photoelectric emission ke liye kaun-kaunsi visible light (400–700 nm) responsible hogi?
Threshold frequency: work function ko Joule me convert karke ν0 = φ0/h se milti hai.
φ0 = 2.14 eV = 2.14 × 1.6×10⁻¹⁹ J = 3.424×10⁻¹⁹ J
ν0 = φ0/h = 3.424×10⁻¹⁹ / 6.63×10⁻³⁴ = 5.16×10¹⁴ Hz
Q2. Q1 wale sodium ke liye threshold wavelength nikalo. Kya poori visible spectrum (400–700 nm) is metal se photoelectric effect de sakti hai?
λ0 = c/ν0 se threshold wavelength milti hai; usse chhoti wavelength (zyada frequency) hi emission karayegi.
λ0 = c/ν0 = (3×10⁸)/(5.16×10¹⁴) ≈ 5.81×10⁻⁷ m = 581 nm
Visible range 400–700 nm me se sirf λ < 581 nm (yani violet-green side) hi photoemission degi; 581–700 nm wali light emission nahi karegi.
Q3. Ek metal ka work function 4.2 eV hai. Us par 1.5×10¹⁵ Hz frequency ki light padti hai. Photoelectrons ki maximum kinetic energy nikalo (Joule aur eV dono me).
Einstein's photoelectric equation KE_max = hν − φ0 use karo — dono terms ko Joule me lao.
φ0 = 4.2 × 1.6×10⁻¹⁹ = 6.72×10⁻¹⁹ J
hν = 6.63×10⁻³⁴ × 1.5×10¹⁵ = 9.945×10⁻¹⁹ J
KE_max = 9.945×10⁻¹⁹ − 6.72×10⁻¹⁹ = 3.225×10⁻¹⁹ J
KE_max (eV) = 3.225×10⁻¹⁹ / 1.6×10⁻¹⁹ ≈ 2.02 eV
Q4. Q3 wale case me stopping potential kitna hoga?
Stopping potential seedha eV0 = KE_max se milta hai, isliye V0 numerically KE_max (eV me) ke barabar hota hai.
eV0 = KE_max = 3.225×10⁻¹⁹ J
V0 = KE_max/e = 3.225×10⁻¹⁹ / 1.6×10⁻¹⁹ ≈ 2.02 V
Q5. λ = 500 nm wavelength ki light ka ek photon lo. Iski energy Joule aur eV me, aur momentum SI units me nikalo.
Photon energy E=hc/λ aur momentum p=h/λ direct formulas hain.
E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(500×10⁻⁹) = 3.978×10⁻¹⁹ J
E (eV) = 3.978×10⁻¹⁹/1.6×10⁻¹⁹ ≈ 2.49 eV
p = h/λ = 6.63×10⁻³⁴/(500×10⁻⁹) = 1.326×10⁻²⁷ kg·m/s
Q6. Ek metal ka threshold wavelength 6800 Å hai. Iska work function eV me nikalo.
φ0 = hc/λ0 use karo, phir Joule se eV me convert karo.
λ0 = 6800 Å = 6.8×10⁻⁷ m
φ0 = hc/λ0 = (6.63×10⁻³⁴ × 3×10⁸)/(6.8×10⁻⁷) = 2.925×10⁻¹⁹ J
φ0 (eV) = 2.925×10⁻¹⁹/1.6×10⁻¹⁹ ≈ 1.83 eV
Q7. Ek electron 100 eV se accelerate hoke chal raha hai. Iski de Broglie wavelength nikalo.
Pehle KE ko Joule me convert karo, phir momentum p=√(2mKE) nikalo, phir λ=h/p.
KE = 100 eV = 100 × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁷ J
p = √(2 × m_e × KE) = √(2 × 9.1×10⁻³¹ × 1.6×10⁻¹⁷) = √(2.912×10⁻⁴⁷) = 5.40×10⁻²⁴ kg·m/s
λ = h/p = 6.63×10⁻³⁴/5.40×10⁻²⁴ ≈ 1.23×10⁻¹⁰ m = 0.123 nm
Q8. Ek electron 1.0×10⁶ m/s speed se move kar raha hai. Iski de Broglie wavelength nikalo.
Yahan velocity direct di hai isliye p=mv seedha use kar sakte hain, KE nikalne ki zaroorat nahi.
p = m_e v = 9.1×10⁻³¹ × 1.0×10⁶ = 9.1×10⁻²⁵ kg·m/s
λ = h/p = 6.63×10⁻³⁴/9.1×10⁻²⁵ ≈ 7.29×10⁻¹⁰ m = 0.729 nm
Q9. Ek electron 150 V potential difference se accelerate hota hai (starting from rest). De Broglie wavelength nikalo.
Accelerating potential se kinetic energy milti hai: KE = eV. Fir p=√(2mKE), phir λ=h/p.
KE = eV = 1.6×10⁻¹⁹ × 150 = 2.4×10⁻¹⁷ J
p = √(2 × 9.1×10⁻³¹ × 2.4×10⁻¹⁷) = √(4.368×10⁻⁴⁷) = 6.61×10⁻²⁴ kg·m/s
λ = h/p = 6.63×10⁻³⁴/6.61×10⁻²⁴ ≈ 1.00×10⁻¹⁰ m = 0.100 nm
Q10. Ek 0.05 kg ki ball 20 m/s speed se chal rahi hai. Iski de Broglie wavelength nikalo aur bata do kyun ye ball 'wave' jaisi behave nahi karti dikhti.
Same formula λ=h/(mv) macroscopic object pe bhi lagta hai, par mass bahut zyada hone se λ negligible ho jaati hai.
p = mv = 0.05 × 20 = 1.0 kg·m/s
λ = h/p = 6.63×10⁻³⁴/1.0 = 6.63×10⁻³⁴ m
Ye wavelength ball ke size (~cm) se 10³⁰ guna chhoti hai, isliye wave nature detect nahi ho pati — sirf atomic-scale particles (electron, proton) me hi ye significant hai.
Q11. Kisi metal surface pe photoelectric effect ke liye threshold frequency 5.0×10¹⁴ Hz hai. Is metal par 8.0×10¹⁴ Hz frequency ki light daali jaati hai. Stopping potential nikalo.
Pehle KE_max = h(ν−ν0) nikalo, phir V0 = KE_max/e.
KE_max = h(ν − ν0) = 6.63×10⁻³⁴ × (8.0×10¹⁴ − 5.0×10¹⁴) = 6.63×10⁻³⁴ × 3.0×10¹⁴ = 1.989×10⁻¹⁹ J
V0 = KE_max/e = 1.989×10⁻¹⁹/1.6×10⁻¹⁹ ≈ 1.24 V
Q12. Ek photon ki energy 3.3 eV hai. Iski wavelength nikalo aur bata do ye electromagnetic spectrum ke kis region me aata hai (approx).
E=hc/λ ko rearrange karke λ=hc/E nikalo, energy ko Joule me convert karke.
E = 3.3 eV = 3.3 × 1.6×10⁻¹⁹ = 5.28×10⁻¹⁹ J
λ = hc/E = (6.63×10⁻³⁴ × 3×10⁸)/5.28×10⁻¹⁹ ≈ 3.77×10⁻⁷ m = 377 nm
Ye visible spectrum ke violet/near-UV region ke paas aata hai.
Q13. Do photons ki wavelengths 400 nm aur 800 nm hain. In dono photons ki energies ka ratio nikalo.
E ∝ 1/λ hai (kyunki E=hc/λ), isliye ratio seedha wavelengths ke inverse ratio se milta hai — full calculation ki zaroorat nahi.
E1/E2 = λ2/λ1 = 800/400 = 2
Yani 400 nm photon ki energy, 800 nm photon se dugni hai (E1 = 2E2).
Q14. Ek proton (mass 1.67×10⁻²⁷ kg) 1000 m/s speed se chal raha hai. Iski de Broglie wavelength nikalo aur electron (Q8, v=10⁶ m/s) ki wavelength se compare karo.
Same λ=h/(mv) formula, par yahan proton ka mass electron se ~1836 guna zyada hai — is baat ka dhyan rakho ki formula sirf mass-velocity pe depend karta hai, particle type pe nahi.
p = mv = 1.67×10⁻²⁷ × 1000 = 1.67×10⁻²⁴ kg·m/s
λ = h/p = 6.63×10⁻³⁴/1.67×10⁻²⁴ ≈ 3.97×10⁻¹⁰ m = 0.397 nm
Compare: Q8 ka electron (v=10⁶ m/s) λ ≈ 0.729 nm tha — proton yahan bhaari hone ke bawajood chhoti speed ki wajah se comparable order ki wavelength de raha hai.
Q15. Ek metal ka work function 2.0 eV hai. Kya isse 700 nm (red light) se photoelectric emission ho sakta hai? Calculation se justify karo.
Photon energy nikalo aur work function se compare karo — agar photon energy kam hai to emission bilkul nahi hoga, intensity badhane se bhi nahi.
E(photon) = hc/λ = (6.63×10⁻³⁴ × 3×10⁸)/(700×10⁻⁹) = 2.84×10⁻¹⁹ J = 1.78 eV
φ0 = 2.0 eV > 1.78 eV
Chunki photon energy work function se kam hai, is red light se emission NAHI hoga — chahe intensity kitni bhi badha do.
Q16. Ek photoelectric experiment me stopping potential 1.5 V mapa gaya. Photoelectrons ki maximum kinetic energy Joule me nikalo.
Direct relation eV0 = KE_max use karo.
KE_max = eV0 = 1.6×10⁻¹⁹ × 1.5 = 2.4×10⁻¹⁹ J
Q17. Kisi metal ke liye ν0 = 6×10¹⁴ Hz hai. Is metal par 9×10¹⁴ Hz light daalne se nikle photoelectron ki de Broglie wavelength nikalo.
Do-step problem: pehle Einstein's equation se KE_max nikalo (ye electron ki KE hai), phir usi KE se de Broglie wavelength nikalo.
KE_max = h(ν−ν0) = 6.63×10⁻³⁴ × (9×10¹⁴ − 6×10¹⁴) = 6.63×10⁻³⁴ × 3×10¹⁴ = 1.989×10⁻¹⁹ J
p = √(2 m_e KE_max) = √(2 × 9.1×10⁻³¹ × 1.989×10⁻¹⁹) = √(3.62×10⁻⁴⁹) = 6.02×10⁻²⁵ kg·m/s
λ = h/p = 6.63×10⁻³⁴/6.02×10⁻²⁵ ≈ 1.10×10⁻⁹ m = 1.10 nm
Q18. Ek photon ka momentum 2×10⁻²⁷ kg·m/s hai. Iski wavelength aur energy (eV me) nikalo.
Pehle λ=h/p se wavelength nikalo, phir E=hc/λ (ya seedha E=pc) se energy.
λ = h/p = 6.63×10⁻³⁴/2×10⁻²⁷ ≈ 3.32×10⁻⁷ m = 332 nm
E = pc = 2×10⁻²⁷ × 3×10⁸ = 6.0×10⁻¹⁹ J
E (eV) = 6.0×10⁻¹⁹/1.6×10⁻¹⁹ ≈ 3.75 eV
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Einstein's photoelectric equation | KEmax = hν − φ0 = hν − hν0 |
| Threshold frequency | ν0 = φ0/h |
| Threshold wavelength | λ0 = hc/φ0 = c/ν0 |
| Stopping potential | eV0 = KEmax |
| Photon energy | E = hν = hc/λ |
| Photon momentum | p = E/c = h/λ |
| de Broglie wavelength (general) | λ = h/p = h/(mv) |
| de Broglie wavelength (from KE) | p = √(2mKE), phir λ = h/p |
| de Broglie wavelength (accelerated charge through V) | KE = qV, phir p = √(2mqV), λ = h/p |
| Constants | h = 6.63×10⁻³⁴ J·s, c = 3×10⁸ m/s, e = 1.6×10⁻¹⁹ C, m_e = 9.1×10⁻³¹ kg, 1 eV = 1.6×10⁻¹⁹ J |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- eV ko Joule me convert karna bhool jaana. h ka unit J·s hai, isliye ν ke saath multiply karke jo energy milti hai wo Joule me hoti hai — agar work function eV me diya hai to use pehle ×1.6×10⁻¹⁹ karke Joule me lao, warna subtraction galat aayega.
- Work function aur threshold frequency ko confuse karna. φ0 energy hai (eV/J), ν0 frequency hai (Hz) — dono related hain (φ0 = hν0) par same cheez nahi. Question me jo diya hai wahi directly formula me daalo, dusre ko usse convert karke nikalo.
- de Broglie wavelength nikalte time KE aur velocity ke formula mix karna. Agar velocity di hai to seedha p=mv use karo. Agar kinetic energy di hai to pehle p=√(2mKE) nikalo — seedha λ=h/(m×KE) jaisa galat formula mat banao.
- Stopping potential ke sign/magnitude me confusion. V0 hamesha ek magnitude hai jo eV0=KE_max se milta hai — negative sign sirf ye batata hai ki retarding potential hai, calculation me numerically KE_max/e hi likho.
- Intensity ko photoelectron ki KE se jodna. Intensity sirf photoelectric CURRENT (electrons ki number) badhati hai, KE_max sirf frequency pe depend karti hai — ye galti bahut common hai kyunki wave theory ka intuition yahi kehta hai (jo galat hai).
- Accelerating potential se seedha wavelength nikalne ki koshish karna. V diya ho to pehle KE=qV se kinetic energy nikalo, phir momentum, phir λ — beech ka step skip karke seedha formula yaad rakhne ki koshish mat karo, galat units aa jaate hain.
Board-Style Important Questions
- 1 mark: Photoelectric effect me stopping potential kis cheez pe depend karta hai — intensity ya frequency?
- 2 marks: Einstein's photoelectric equation likho aur ismein har term ka matlab batao.
- 2 marks: De Broglie relation likho aur ismein h aur p ka matlab batao.
- 3 marks: Ek metal ka work function diya ho aur usspar padne wali light ki frequency di ho, to maximum kinetic energy aur stopping potential nikalne wala numerical solve karo.
- 3 marks: Threshold frequency se kam frequency wali light se photoelectric emission kyun nahi hota — wave theory se compare karke Einstein ke explanation ke through samjhao.
- 5 marks: Photoelectric effect ki saari experimental observations (laws) likho, aur Einstein's photon theory in sabko kaise explain karti hai wo detail me batao.
Aksar Poochhe Jaane Wale Sawaal
Photoelectric effect aur photon concept ka kya relation hai?
Photoelectric effect ke experimental observations (jaise threshold frequency, koi time lag na hona) wave theory se explain nahi hoti thi. Einstein ne propose kiya ki light photons (discrete energy packets, E=hν) ki stream hai — is photon model se saari observations explain ho gayin, isliye photoelectric effect photon concept ka strongest experimental proof hai.
Work function aur threshold frequency me kya difference hai?
Work function (φ0) wo minimum energy hai jo electron ko metal surface se bahar nikalne ke liye chahiye (unit: eV ya J). Threshold frequency (ν0) wo minimum frequency hai jispe ye energy exactly match hoti hai (φ0=hν0). Dono ek hi cheez ko do units me represent karte hain — ek energy me, doosra frequency me.
Intensity badhane se photoelectron ki kinetic energy kyun nahi badhti?
Kyunki har photoelectron sirf ek hi photon se energy leta hai (one-to-one interaction). Intensity badhane ka matlab hai zyada photons per second — isse zyada electrons emit honge (current badhega), par har electron ko milne wali energy (hν) same rahegi, isliye KE_max same rahegi.
De Broglie wavelength sirf electron ke liye hi applicable hai kya?
Nahi, λ=h/(mv) formula har moving particle ke liye applicable hai — electron, proton, neutron, ya ek cricket ball bhi. Farak sirf itna hai ki bade (macroscopic) objects ki wavelength itni chhoti hoti hai (10⁻³⁴ m order) ki wo practically detect nahi ho pati; sirf atomic-scale particles me ye significant hoti hai.
Photon ka rest mass zero hone ke bawajood momentum kaise hota hai?
Photon hamesha speed of light c se travel karta hai, isliye classical formula p=mv apply nahi hota. Iske bajaye relativistic relation E=pc use hota hai, jisse p=E/c=h/λ milta hai — ye momentum wave nature (λ) se directly connected hai.
Board exam me is chapter ke numericals me sabse zyada kya galti hoti hai?
Sabse common galti hai unit conversion — eV aur Joule ke beech confuse ho jaana, ya kinetic energy aur velocity wale de Broglie formulas ko mix kar dena. Har numerical me sabse pehle check karo ki diya gaya data kis unit me hai, aur formula me daalne se pehle sabko consistent SI units me convert karo.
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