Photoelectric effect — Einstein's explanation, work function: Worked examples

2.3.2 · D3 · Physics › Modern Physics › Photoelectric effect — Einstein's explanation, work function

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Scenario matrix

Har photoelectric problem inhi cells mein se ek hota hai. Neeche ke examples cell letter se tagged hain, taaki milake sab cover ho jaayein. Har cell figure mein $V_{0}$ -vs- $ν$ line par ek jagah bhi live karti hai — table aur figure saath padhein.

#	Cell (case class)	Kya diya → kya poochha	Example
A	Standard forward	$λ$ (ya $ν$ ), $ϕ$ → $K E_{m a x}$ , $V_{0}$	Ex 1
B	Threshold / degenerate	$ϕ$ → $λ_{0}$ ya $ν_{0}$ ("just barely" edge, $K E_{m a x} = 0$ )	Ex 2
C	Below threshold (zero output)	$h ν < ϕ$ → prove karo kuch emit nahi hota	Ex 3
D	Reverse solve for $ϕ$	$λ$ , $V_{0}$ → $ϕ$	Ex 4
E	Two-point slope (measure $h$ )	$(ν_{1}, V_{0, 1}), (ν_{2}, V_{0, 2})$ → $h$ , $ϕ$	Ex 5
F	Ratio / scaling	$λ$ aadha karo, $K E_{m a x}$ par kya asar? (non-proportional trap)	Ex 6
G	Intensity vs frequency (word problem)	brightness double karo — kya $V_{0}$ badlega?	Ex 7
H	Limiting behaviour	$ν \to \infty$ aur $ν \to ν_{0}^{+}$ — line ki shape	Ex 8
I	Exam twist (electron speed)	$K E_{m a x}$ → actual velocity $v_{m a x}$ m/s mein	Ex 9

Neeche ka figure dekho — poora matrix ek straight-line graph of $V_{0}$ vs $ν$ par live karta hai. Lavender line hai $V_{0} = (h / e) ν - ϕ / e$ ; iska slope (steepness) hai $h / e$ , iska $x$ -intercept (red dot on the axis) threshold $ν_{0}$ hai, aur iska $y$ -intercept hai $- ϕ / e$ . Mint-shaded strip left mein Cell C hai, jahan koi line exist nahi karti kyunki koi electron escape nahi karta. Har labelled dot ek cell letter ko uski geometric jagah se jodata hai.

Figure — Photoelectric effect — Einstein's explanation, work function

Examples

Ex 1 — Cell A: standard forward

Worked example Sodium par violet light

$ϕ = 2.3$ eV, wavelength $λ = 400$ nm ki light. $K E_{m a x}$ aur $V_{0}$ nikalo.

Forecast: solve karne se pehle andaza lagao — kya electron tez niklega ya mushkil se? (Violet kaafi energetic hai; $ϕ$ modest hai, toh ek chota sa positive KE expect karo.) Yeh figure mein grey dot hai, line par threshold ke upar baiitha hai.

Step 1. Photon energy: $E = \frac{1240}{400} = 3.10$ eV. Yeh step kyun? Ek packet ki value jaanno zaroori hai tab subtract karo toll.

Step 2. $K E_{m a x} = E - ϕ = 3.10 - 2.3 = 0.80$ eV. Yeh step kyun? Einstein's equation — $ϕ$ pay karne ke baad bacha energy.

Step 3. $V_{0} = K E_{m a x} / e = 0.80$ V. Full SI mein: $K E_{m a x} = 0.80 eV = 0.80 \times 1.6 \times 1 0^{- 19} = 1.28 \times 1 0^{- 19}$ J, aur $V_{0} = \frac{K E _{m a x}}{e} = \frac{1.28 \times 1 0 ^{- 19} J}{1.6 \times 1 0 ^{- 19} C} = 0.80$ V. Yeh step kyun? Joules-and-coulombs division ko poori tarah karne se dikhta hai kyun eV number aur volt number identical aate hain — $1.6 \times 1 0^{- 19}$ ke dono factors cancel ho jaate hain.

Verify: SI division $0.80$ V deta hai, eV shortcut se match karta hai ✓. Sanity check: $K E_{m a x} = 0.80$ eV positive hai aur photon energy se kam hai, jaisa hona chahiye (jitna mila usse zyada nahi rakeh sakte). ✓

Ex 2 — Cell B: threshold (degenerate edge)

Worked example Woh sabse lambi wavelength jo abhi bhi kaam kare

Ek metal ka $ϕ = 4.0$ eV hai. Woh sabse lambi wavelength $λ_{0}$ kya hai jo electrons eject karti hai?

Forecast: "Sabse lambi wavelength" = sabse kam energy = just-barely case. Expect karo $K E_{m a x} = 0$ answer par, aur wavelength UV mein honi chahiye (kyunki 4 eV kisi bhi visible photon se zyada hai). Figure mein yeh red dot on the $ν$ -axis hai jahan line touch karti hai.

Step 1. Threshold par, $K E_{m a x} = 0$ , isliye poori photon energy escape mein kharch hoti hai: $ϕ = \frac{h c}{λ _{0}}$ . Yeh step kyun? Lambi wavelength = kamzor photon; $λ$ tab tak badhao jab tak photon ke paas bilkul utna hi ho aur kuch bachaa nahi. Yahi emission ki boundary hai.

Step 2. $λ_{0} = \frac{1240}{4.0} = 310$ nm. Yeh step kyun? eV–nm shortcut ko $λ$ ke liye rearrange karo.

Verify: $310$ nm ultraviolet hai (visible light ki $\sim 400$ nm violet edge ke neeche) ✓ — isliye is metal par ordinary visible light fail karti hai, bilkul wahi jo "threshold" idea predict karta hai. Plug back: $1240/310 = 4.0$ eV $= ϕ$ ✓.

Ex 3 — Cell C: below threshold (zero output)

Worked example Usi metal par red light

Wahi metal, $ϕ = 4.0$ eV, ab $λ = 620$ nm (red) se lit hai. $K E_{m a x}$ kya hai? Ek electron free karne ke liye kaun kitne red photons per second chahiye?

Forecast: Red kam energy wali hai. $310$ nm threshold se compare karein, $620$ nm lambi hai → kamzor photon → shayad kuch nahi niklega. Yeh figure mein mint-shaded strip hai — $ν_{0}$ ke baayein, koi line hi nahi. Aur "kitne photons" waala question ek trap hai.

Step 1. Photon energy: $E = \frac{1240}{620} = 2.0$ eV. Yeh step kyun? Ek packet ko toll se compare karte hain.

Step 2. $E - ϕ = 2.0 - 4.0 = - 2.0$ eV. Negative "leftover" impossible hai — physically matlab hai koi electron escape nahi karta, isliye $K E_{m a x} = 0$ (koi ejected electrons hain hi nahi). Yeh step kyun? Ek photon ek electron ke liye pay karta hai; agar single packet toll cover nahi kar sakta, electron wahi rahega.

Step 3. Kitne photons chahiye: irrelevant — koi bhi number kaam nahi karta. Electrons instantly de-excite ho jaate hain aur bahut saare weak photons "save up" nahi kar sakte. Yeh step kyun? Yeh classical accumulation myth hai; har absorption ek one-shot event hai.

Verify: $2.0 eV < 4.0 eV = ϕ$ , isliye $h ν < ϕ$ ✓ → sub-threshold case, emission = zero regardless of intensity. Ex 2 se consistent: $620 > 310$ nm matlab below threshold. ✓

Ex 4 — Cell D: reverse-solve for $ϕ$

Worked example Stopping voltage diya hai, metal ka work function nikalo

$λ = 300$ nm ki light ek unknown metal par, stopping potential $V_{0} = 1.13$ V deta hai. $ϕ$ nikalo.

Forecast: Hum equation ulta chalayenge: photon energy minus measured KE = toll. Expect karo $ϕ$ around 3 eV.

Step 1. Photon energy: $E = \frac{1240}{300} = 4.133$ eV. Yeh step kyun? Pehle packet ki value hamesha establish karo.

Step 2. $K E_{m a x} = e V_{0} = 1.13$ eV (volts → eV directly). Yeh step kyun? Stopping potential measure karta hai fastest electron ka KE.

Step 3. $ϕ = E - K E_{m a x} = 4.133 - 1.13 = 3.00$ eV. Yeh step kyun? Rearranged Einstein equation: toll = packet − leftover.

Verify: Forward-check: $K E_{m a x} = 4.133 - 3.00 = 1.13$ eV → $V_{0} = 1.13$ V ✓ given se match karta hai. Is metal ka threshold: $λ_{0} = 1240/3.00 = 413$ nm (violet), aur $300 < 413$ nm isliye emission expected hai ✓.

Ex 5 — Cell E: do data points → $h$ measure karo (Millikan style)

Worked example Voltage graph se Planck's constant nikalo

Ek metal ke liye: $ν_{1} = 7 \times 1 0^{14}$ Hz par $V_{0} = 0.50$ V; $ν_{2} = 10 \times 1 0^{14}$ Hz par $V_{0} = 1.74$ V. $h$ aur $ϕ$ nikalo.

Forecast: Line $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ par do points hain. Slope $h / e$ deta hai; $V_{0} = 0$ tak extend karo toh $ν_{0}$ milega, isse $ϕ$ . Expect karo $h \approx 6.6 \times 1 0^{- 34}$ . Yeh figure mein do butter-yellow squares hain.

Step 1. Slope $= \frac{Δ V _{0}}{Δ ν} = \frac{1.74 - 0.50}{( 10 - 7 ) \times 1 0 ^{14}} = 4.133 \times 1 0^{- 15}$ V·s. Yeh step kyun? $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ se, $ν$ ka coefficient hai $h / e$ — straight line ki geometry $h$ encode karti hai.

Step 2. $h = slope \times e = 4.133 \times 1 0^{- 15} \times 1.6 \times 1 0^{- 19} = 6.61 \times 1 0^{- 34}$ J·s. Yeh step kyun? $/ e$ multiply karo $h$ isolate karne ke liye.

Step 3. Point 1 use karke $ϕ$ nikalo: $e V_{0, 1} = h ν_{1} - ϕ$ , isliye $ϕ = h ν_{1} - e V_{0, 1} = (6.61 \times 1 0^{- 34}) (7 \times 1 0^{14}) - (1.6 \times 1 0^{- 19}) (0.50) .$ Yeh hai $ϕ = 4.63 \times 1 0^{- 19} - 0.80 \times 1 0^{- 19} = 3.83 \times 1 0^{- 19}$ J $= 2.39$ eV. Yeh step kyun? Ek baar slope pata ho, koi bhi ek point intercept fix kar deta hai, yaani toll.

Verify: $h = 6.61 \times 1 0^{- 34}$ J·s accepted $6.63 \times 1 0^{- 34}$ se rounding ke andar match karta hai ✓. Point 2 ke saath cross-check $h ν_{2}$ directly compute karke: $h ν_{2} = (6.61 \times 1 0^{- 34}) (10 \times 1 0^{14}) = 6.61 \times 1 0^{- 19}$ J. Phir $K E_{m a x} = h ν_{2} - ϕ = 6.61 \times 1 0^{- 19} - 3.83 \times 1 0^{- 19} = 2.78 \times 1 0^{- 19}$ J $= 1.74$ eV → $V_{0} = 1.74$ V ✓, exactly diya gaya second point.

Ex 6 — Cell F: ratio / scaling trap

Worked example Wavelength aadhi karo — kya

$K E_{m a x}$ double hota hai? Ek metal ka $ϕ = 2.0$ eV hai. $λ_{1} = 400$ nm ki light kuch $K E_{m a x}$ deti hai. Agar wavelength aadhi karke $λ_{2} = 200$ nm kar dein, toh kya $K E_{m a x}$ double hoga?

Forecast: Dhyan raho! Jab wavelength aadhi hoti hai toh photon energy double hoti hai, lekin $K E_{m a x}$ mein $ϕ$ subtract hota hai, isliye yeh double se zyada hoga. Common trap: log "just doubles" answer karte hain.

Step 1. $E_{1} = 1240/400 = 3.10$ eV → $K E_{m a x, 1} = 3.10 - 2.0 = 1.10$ eV. Yeh step kyun? Baseline value.

Step 2. $E_{2} = 1240/200 = 6.20$ eV → $K E_{m a x, 2} = 6.20 - 2.0 = 4.20$ eV. Yeh step kyun? $λ$ aadha karne se photon energy double hoti hai ( $6.20 = 2 \times 3.10$ ✓), lekin toll $ϕ$ fixed subtraction hai.

Step 3. Ratio $K E_{m a x, 2} / K E_{m a x, 1} = 4.20/1.10 = 3.82$ , nahi 2. Yeh step kyun? Kyunki $K E_{m a x} = E - ϕ$ affine hai, proportional nahi — ek constant subtract karna simple doubling tod deta hai.

Verify: Photon energies did double ( $3.10 \to 6.20$ ) ✓. Phir bhi KE factor $3.82 > 2$ se badha ✓, trap confirm karta hai. Sirf photon energy $1/ λ$ se scale karti hai, $K E_{m a x}$ nahi.

Ex 7 — Cell G: intensity vs frequency (word problem)

Worked example Brightness waala question

Tum caesium ( $ϕ = 2.1$ eV) par $λ = 350$ nm ki light ( $>$ threshold) daalo aur $V_{0}$ record karo. Ek dost brightness double karta hai (same colour). Kya $V_{0}$ badlega? Kya photocurrent badlega?

Forecast: Brightness = photons per second, energy per photon nahi. Predict karo: $V_{0}$ nahi badlega, current double hoga.

Step 1. Photon energy: $E = 1240/350 = 3.543$ eV. Yeh step kyun? Per-packet energy establish karo — yahi $K E_{m a x}$ set karta hai.

Step 2. $K E_{m a x} = 3.543 - 2.1 = 1.44$ eV → $V_{0} = 1.44$ V. Intensity double karne se $E$ , $ϕ$ , ya $K E_{m a x}$ nahi badlta, isliye $V_{0}$ $1.44$ V hi rahega. Yeh step kyun? Har electron bilkul ek photon "khata" hai; photon ki energy colour ( $ν$ ) par depend karti hai, kitne aate hain us par nahi.

Step 3. Photocurrent double hoga: double photons per second → double ejected electrons per second. Yeh step kyun? Current electrons count karta hai, aur electron count photon count track karta hai.

Verify: $E = 3.543 > ϕ = 2.1$ isliye emission hoti hai ✓. $V_{0}$ sirf $(ν, ϕ)$ par depend karta hai, dono nahi badla, isliye $V_{0} = 1.44$ V pehle aur baad mein ✓ — parent note ke classical-vs-quantum table se match karta hai.

Ex 8 — Cell H: line ka limiting behaviour

Worked example Graph ke do ends

$ϕ = 2.0$ eV wale metal ke $V_{0}$ vs $ν$ line par: (a) $ν \to ν_{0}^{+}$ (threshold ke just upar) par $V_{0}$ kya hoga? (b) $ν \to \infty$ par $V_{0}$ kaise behave karta hai?

Forecast: Low end par $V_{0} \to 0$ (electrons mushkil se escape karte hain). High end par $V_{0}$ linearly aur unboundedly badhta jaata hai. Figure ke two flagged ends dekho — red dot (low) aur lavender arrow (high).

Step 1. Work function pehle joules mein convert karo: $ϕ = 2.0 eV \times 1.6 \times 1 0^{- 19} J/eV = 3.2 \times 1 0^{- 19}$ J. Phir threshold: $ν_{0} = \frac{ϕ}{h} = \frac{3.2 \times 1 0 ^{- 19} J}{6.63 \times 1 0 ^{- 34} J\cdots} = 4.83 \times 1 0^{14}$ Hz. Yeh step kyun? $ν_{0}$ woh jagah hai jahan line $ν$ -axis se milti hai; uske neeche koi line hai hi nahi (koi emission nahi). Hume eV→J convert karna padta hai kyunki $h$ SI units mein hai.

Step 2. $ν \to ν_{0}^{+}$ par: $K E_{m a x} = h ν - ϕ \to 0$ , isliye $V_{0} \to 0$ . Yeh step kyun? Threshold par photon ka poora hissa toll bharta hai; rokne ke liye kuch bachi nahi.

Step 3. $ν \to \infty$ par: $V_{0} = \frac{h}{e} ν - \frac{ϕ}{e}$ linearly aur unboundedly badhta hai; slope hamesha $h / e = 4.14 \times 1 0^{- 15}$ V·s rehta hai. Yeh step kyun? $ϕ$ term ek huge $h ν$ ke aage negligible ho jaati hai; line ki steepness hamesha ke liye fixed hai.

Verify: $ν_{0} = 4.83 \times 1 0^{14}$ Hz correspond karta hai $λ_{0} = c / ν_{0} = 3 \times 1 0^{8} /4.83 \times 1 0^{14} = 621$ nm ✓ aur shortcut se $λ_{0} = 1240/2.0 = 620$ nm ✓ (rounding ke andar agree karte hain). Slope $h / e = 6.63 \times 1 0^{- 34} /1.6 \times 1 0^{- 19} = 4.14 \times 1 0^{- 15}$ V·s ✓.

Ex 9 — Cell I: exam twist (actual electron speed)

Worked example Energy se velocity tak

Ek electron $K E_{m a x} = 0.80$ eV ke saath metal chhod kar jaata hai. Uski maximum speed $v_{m a x}$ nikalo. (Electron mass $m_{e} = 9.11 \times 1 0^{- 31}$ kg.)

Forecast: Pehle eV ko joules mein convert karo, phir $K E = \frac{1}{2} m v^{2}$ ko invert karo. Expect karo $v$ around $5 \times 1 0^{5}$ m/s (electrons halke hote hain, isliye thodi si energy bhi high speed deti hai).

Step 1. Energy SI mein convert karo: $K E_{m a x} = 0.80 \times 1.6 \times 1 0^{- 19} = 1.28 \times 1 0^{- 19}$ J. Yeh step kyun? Kinetic-energy formula ko joules, kilograms, metres chahiye — eV mix nahi kar sakte.

Step 2. $K E = \frac{1}{2} m v^{2}$ se: $v_{m a x} = \frac{2 K E _{m a x}}{m _{e}} = \frac{2 ( 1.28 \times 1 0 ^{- 19} )}{9.11 \times 1 0 ^{- 31}}$ . Yeh step kyun? Hume energy aur mass pata hai aur speed chahiye — classical KE definition invert karo (non-relativistic kyunki $v ≪ c$ ).

Step 3. $v_{m a x} = 2.81 \times 1 0^{11} = 5.30 \times 1 0^{5}$ m/s. Yeh step kyun? Numeric ratio ka square root lo.

Verify: $v_{m a x} = 5.3 \times 1 0^{5}$ m/s $\approx 0.0018 c$ hai, safely non-relativistic ✓ isliye classical formula valid thi. Forward-check: $\frac{1}{2} (9.11 \times 1 0^{- 31}) (5.30 \times 1 0^{5})^{2} = 1.28 \times 1 0^{- 19}$ J $= 0.80$ eV ✓.

Recall Yeh problem kaun sa cell hai?

Wavelength aur work function diya, stopping potential poochha? ::: Cell A — standard forward (Ex 1). Sirf $ϕ$ diya, sabse lambi wavelength poochhi? ::: Cell B — threshold, $K E_{m a x} = 0$ set karo (Ex 2). Photon energy $ϕ$ se kam hai? ::: Cell C — koi emission nahi, kisi bhi intensity par (Ex 3). $λ$ aur $V_{0}$ diya, $ϕ$ poochha? ::: Cell D — reverse solve (Ex 4). Do $(ν, V_{0})$ pairs, $h$ poochha? ::: Cell E — slope $= h / e$ (Ex 5). Kya $λ$ aadha karne se $K E_{m a x}$ double hota hai? ::: Nahi — Cell F, KE affine hai proportional nahi (Ex 6). Kya brightness double karne se $V_{0}$ badlta hai? ::: Nahi — Cell G, sirf current badlta hai (Ex 7).

Connections

Parent: Einstein's photoelectric equation — yeh page uska worked-example drill hai.
Energy conservation — har example sirf $h ν = ϕ + K E_{m a x}$ ki bookkeeping hai.
Millikan oil drop & measurement of h — Ex 5 woh slope method hai jisne $h$ measure kiya.
Planck's quantum hypothesis & blackbody radiation — jahan se $E = h ν$ aata hai.
Wave-particle duality of light — kyun "one photon, one electron" legitimate hai.
de Broglie wavelength — Ex 9 ka electron speed matter waves mein le jaata hai.
Compton effect — photon-as-particle ka aur evidence.