Photon properties — E = hf, p = h - λ: Worked examples

2.3.3 · D3 · Physics › Modern Physics › Photon properties — E = hf, p = h - λ

Yeh page parent photon note ka practice engine hai. Wahan humne do formulas banayi thin. Yahan hum unhe har tarah ki situation se guzarte hain jo exam ya real problem mein aa sakti hai — taaki jab bhi naya question dekho, tum turant pehchaan lo ki woh table ke kis cell mein aata hai.

Constants jo poore note mein use honge (inhe yaad karo):

Scenario matrix

Kuch bhi kaam karne se pehle, yahan poora landscape hai. Har row ek class ka problem hai; aakhri column mein us example ka naam hai jo ise cover karta hai. Agar tum yeh sab kar sako, toh koi bhi photon question tumhe surprise nahi kar sakta.

Cell	Kya vary karta hai / twist kya hai	Diya → Chahiye	Covered by
A. $λ$ diya hai	wavelength pata hai	$λ \to E$	Example 1
B. $f$ diya hai	frequency pata hai	$f \to E$	Example 2
C. Momentum route	massless hai, $p$ chahiye	$λ \to p$ , cross-check via $E / c$	Example 3
D. Beam / power	bahut saare photons	$P, λ \to N$ per second	Example 4
E. Scaling (ratio)	numbers nahi, bas "×2"	$λ \to E, p$ proportion	Example 5
F. Limiting / degenerate	$λ \to \infty$ (radio), $λ \to 0$ (gamma), $f = 0$	extremes par behavior	Example 6
G. Matter par push	real world	photon stream se force	Example 7
H. Exam twist	eV diya hai, ya ulta kaam	$E$ (eV) $\to λ$ ; colour identify karo	Example 8

Neeche ki figure mental map hai — har cell ek alag darwaza hai, andar jaate hain usi teen formulas mein.

Figure — Photon properties — E = hf, p = h - λ

Worked Examples

Cell A — tumhe wavelength di gayi hai

Worked example 1 — 400 nm (violet) photon ki energy

Ek photon ki wavelength $λ = 400 nm$ hai. Uski energy joules aur eV mein nikalo.

Forecast: Violet high-frequency hoti hai (short wavelength), toh socho photon energy badi hogi — kuch eV. Yeh guess yaad rakhna.

Wavelength ko metres mein convert karo: $λ = 400 \times 1 0^{- 9} m$ . Yeh step kyun? Har SI formula metres chahta hai; "nm" rakhne se units silently galat ho jaate hain.
Woh form chuno jisme $λ$ ho: $E = \frac{h c}{λ}$ . Yeh step kyun? Humein $λ$ diya hai, $f$ nahi — $h c / λ$ use karne se koi extra conversion nahi lagti.
Plug in karo: $E = \frac{1.986 \times 1 0 ^{- 25}}{400 \times 1 0 ^{- 9}} = 4.97 \times 1 0^{- 19} J$ .
eV mein convert karo: $E = \frac{4.97 \times 1 0 ^{- 19}}{1.602 \times 1 0 ^{- 19}} = 3.10 eV$ . Yeh step kyun? eV photons ke liye human-scale unit hai; $3.10$ eV ko bond/atomic energies se seedha compare kar sakte ho.

Verify: Shortcut use karo $E (eV) = \frac{1240}{λ ( nm )} = \frac{1240}{400} = 3.10 eV$ ✓. "Kuch eV" ka forecast match karta hai.

Cell B — tumhe frequency di gayi hai

Cell C — momentum route (massless!)

Worked example 3 — 500 nm photon ka momentum, do tarike se

$λ = 500 nm$ photon ka momentum nikalo, aur confirm karo ki dono routes agree karte hain.

Forecast: Photon zero rest mass ke bawajood momentum carry karta hai. Yeh bahut chota hoga — around $1 0^{- 27} kg\cdotm/s$ . Exponent guess karo.

Direct route: $p = \frac{h}{λ} = \frac{6.626 \times 1 0 ^{- 34}}{500 \times 1 0 ^{- 9}} = 1.33 \times 1 0^{- 27} kg\cdotm/s$ . Yeh step kyun? $p = h / λ$ ko sirf woh wavelength chahiye jo diya gaya hai.
Energy ke zariye independent route: Example 2 se, $E = 3.98 \times 1 0^{- 19} J$ , aur $p = \frac{E}{c}$ . Yeh step kyun? Massless particle ke liye relativistic relation collapse hokar $E = p c$ ban jaata hai, isliye $p = E / c$ ek alag raasta hai — cross-checking ke liye perfect.
Compute karo: $p = \frac{3.98 \times 1 0 ^{- 19}}{3 \times 1 0 ^{8}} = 1.33 \times 1 0^{- 27} kg\cdotm/s$ .

Verify: Dono routes $1.33 \times 1 0^{- 27} kg\cdotm/s$ dete hain ✓. Yeh agreement hi proof hai ki light ke liye $E = p c$ holds karta hai. Units: $\frac{J\cdots}{m} = \frac{kg\cdotm ^{2} / s ^{2} \cdot s}{m} = kg\cdotm/s$ ✓. Relativistic Energy-Momentum Relation dekho — $E = p c$ kyun hota hai.

Cell D — ek beam: bahut saare photons count karna

Worked example 4 — 5 mW green pointer (

$λ = 532$ nm) se har second kitne photons? Ek laser pointer $P = 5 mW$ power emit karta hai $λ = 532 nm$ par. Har second kitne photons nikalte hain?

Forecast: Har photon ~ $1 0^{- 19}$ J ka hai aur beam har second $1 0^{- 3}$ J pour karti hai, toh around $1 0^{16}$ photons/s expect karo — ek mind-bendingly badi number. Power of ten guess karo.

Har photon ki energy: $E = \frac{1240}{532} = 2.33 eV = 2.33 \times 1.602 \times 1 0^{- 19} = 3.73 \times 1 0^{- 19} J$ . Yeh step kyun? Power "energy per second" hai; isko photon count mein badalne ke liye ek photon ki energy pata honi chahiye.
Photons per second: $N = \frac{P}{E} = \frac{5 \times 1 0 ^{- 3}}{3.73 \times 1 0 ^{- 19}}$ . Yeh step kyun? $P = N \times E$ (count times energy each = total energy per second), isliye $N = P / E$ .
Compute karo: $N = 1.34 \times 1 0^{16}$ photons/s.

Verify: Multiply back karo: $N \times E = 1.34 \times 1 0^{16} \times 3.73 \times 1 0^{- 19} = 5.0 \times 1 0^{- 3} W = 5 mW$ ✓. Isliye ek "weak" beam bhi bilkul continuous dikhti hai — Wave-Particle Duality dekho.

Cell E — pure scaling, koi numbers nahi chahiye

Worked example 5 — Wavelength teen guna karo:

$E$ aur $p$ ka kya hoga? Ek photon ki wavelength badhakar uski original value ki 3 guna kar di jaati hai. Uski energy aur momentum kis factor se change honge?

Forecast: Redder = longer wavelength = kamzor photon. Dono $E$ aur $p$ ghatne chahiye. Factor guess karo.

Dono formulas ko proportions ki tarah likho: $E = \frac{h c}{λ} \propto \frac{1}{λ}$ aur $p = \frac{h}{λ} \propto \frac{1}{λ}$ . Yeh step kyun? Jab sirf ratio puchha ho, constants ko symbolic rakhna chahiye — woh cancel ho jaate hain.
$λ \to 3 λ$ replace karo: new $E = \frac{h c}{3 λ} = \frac{1}{3} E_{old}$ ; $p$ ke saath bhi same. Yeh step kyun? Substituting karne se pata chalta hai ki constant $h c$ (ya $h$ ) intact rehta hai, isliye sirf $1/3$ factor matter karta hai.

Verify: Dono ek tihaai ho jaate hain. Numbers se check karo: $λ = 500 \to 1500 nm$ deta hai $E = 1240/1500 = 0.827 eV$ , aur $2.48/3 = 0.827 eV$ ✓. Yeh inverse- $λ$ behaviour exactly wahi hai jo de Broglie Wavelength ko power karta hai.

Cell F — extreme / degenerate cases

Worked example 6 — Teen limits: radio, gamma, aur "

$f = 0$ " (a) Ek long radio wave $λ = 3 km$ , (b) ek gamma ray $λ = 1 pm = 1 0^{- 12} m$ , aur (c) degenerate case $f \to 0$ ke liye photon energy describe karo.

Forecast: Radio → almost koi energy nahi; gamma → enormous energy; $f = 0$ → bilkul bhi photon nahi. Har ek confirm karo.

(a) Radio, $λ = 3000 m$ : $E = \frac{1240 eV\cdotnm}{3000 \times 1 0 ^{9} nm} = 4.1 \times 1 0^{- 10} eV$ . Yeh step kyun? 3000 m ko nm mein convert karo ( $\times 1 0^{9}$ ) taaki 1240 shortcut reuse ho sake. Energy vanishingly chhoti hai — ek radio photon kuch bhi zyada nahi hilata, isliye radio itna wave-like behave karta hai.
(b) Gamma, $λ = 1 0^{- 12} m = 1 0^{- 3} nm$ : $E = \frac{1240}{1 0 ^{- 3}} = 1.24 \times 1 0^{6} eV = 1.24 MeV$ . Yeh step kyun? Same formula, opposite extreme — ek akela gamma photon MeV energies carry karta hai, itni ki electrons ko atoms se bahar nikal sake (ionising radiation).
(c) $f \to 0$ : $E = h f \to 0$ aur $p = E / c \to 0$ . Koi photon nahi hai — zero-frequency oscillation kuch bhi carry nahi karta. Yeh step kyun? Degenerate limit check karna confirm karta hai ki formula sensibly behave karta hai: koi oscillation nahi, koi quantum nahi.

Verify: Jab $λ$ $3 km$ se $1 pm$ tak jaata hai, $E$ $\sim 1 0^{- 10} eV$ se $\sim 1 0^{6} eV$ tak chadh jaata hai — 15 orders of magnitude, sab ek $1/ λ$ law se ✓. Neeche ki figure yeh plot karti hai. Radio: $1240/ (3000 \times 1 0^{9}) = 4.13 \times 1 0^{- 10}$ eV ✓; Gamma: $1240/1 0^{- 3} = 1.24 \times 1 0^{6}$ eV ✓.

Cell G — real-world push

Worked example 7 — Solar sail par sunlight ki force jab poori absorb ho

$I = 1360 W/m^{2}$ intensity ki sunlight $A = 1000 m^{2}$ area ki sail par girti hai aur poori absorb hoti hai. Sail par kitni force lagti hai?

Forecast: Light push karti hai (comet tails, solar sails). Lekin $p = E / c$ mein $c$ bahut bada hai, isliye force chhhoti hogi — kuch millinewtons. Guess karo.

Sail par padne wali power: $P = I \times A = 1360 \times 1000 = 1.36 \times 1 0^{6} W$ . Yeh step kyun? Intensity power per area hai; total power woh cheez hai jo momentum carry karti hai.
Har second pahunchne wala momentum: light ka har joule $p = E / c$ carry karta hai, isliye momentum per second $= \frac{P}{c}$ . Yeh step kyun? Force hi hai momentum delivered per second (Newton's second law), aur photon momentum $E / c$ hai — Radiation Pressure dekho.
Force (full absorption): $F = \frac{P}{c} = \frac{1.36 \times 1 0 ^{6}}{3 \times 1 0 ^{8}} = 4.53 \times 1 0^{- 3} N$ . Yeh step kyun? Absorb hui light apna poora momentum ek baar de deti hai; ek mirror yeh double kar deta (woh reflect karta hai, momentum reverse hota hai).

Verify: Ek badi sail par $4.53 mN$ — chhota lekin continuous, aur space mein bina friction ke yeh kisi spacecraft ko mahino mein accelerate kar deta hai ✓. Units: $\frac{W}{m/s} = \frac{J/s}{m/s} = \frac{kg\cdotm ^{2} / s ^{3}}{m/s} = kg\cdotm/s^{2} = N$ ✓.

Cell H — exam twist (ulta kaam)

Worked example 8 — Energy di gayi hai, wavelength nikalo aur colour batao

Ek photon exactly $E = 1.77 eV$ carry karta hai. Uski wavelength nikalo aur colour identify karo.

Forecast: 2 eV se kam matlab low-energy visible light → yeh red end mein hona chahiye, around 700 nm. Guess karo.

Shortcut $E (eV) = \frac{1240}{λ ( nm )}$ ko $λ$ ke liye rearrange karo: $λ = \frac{1240}{E ( eV )}$ . Yeh step kyun? Humein $E$ diya hai aur $λ$ chahiye — usi relation ko ulta solve karna $h c$ se re-derive karne se bachata hai.
Plug in karo: $λ = \frac{1240}{1.77} = 700.6 nm$ .
Identify karo: 700 nm deep red hai (visible band ~380–700 nm hai). Yeh step kyun? Colour naam karna woh sanity check hai jo examiner chahta hai — aur yeh Photoelectric Effect intuition se match karta hai ki redder = less energetic.

Verify: Full constants se forward-check karo: $E = \frac{1.986 \times 1 0 ^{- 25}}{700.6 \times 1 0 ^{- 9}} = 2.835 \times 1 0^{- 19} J = 1.77 eV$ ✓. "Red, ~700 nm" ka forecast confirm hua.

Recall Rapid-fire self-test (answers dhako)

400 nm photon ki energy eV mein? ::: $3.10$ eV $2.48$ eV ka photon — wavelength kya? ::: $500$ nm Agar $λ$ teen guna ho, $E$ kya ho jaata hai? ::: ek tihaai 1 pm par gamma ray — rough energy? ::: lagbhag $1.24$ MeV Poori absorb hui power $P$ se force? ::: $F = P / c$ 500 nm photon ka momentum? ::: $1.33 \times 1 0^{- 27}$ kg·m/s

Connections

Photoelectric Effect — Cell H ka "colour vs energy" logic exactly iska threshold idea hai.
Radiation Pressure — Cell G ka macroscopic form.
Relativistic Energy-Momentum Relation — $p = E / c$ ko justify karta hai jo Cell C mein use hua.
de Broglie Wavelength — Cell E ka $1/ λ$ scaling, matter ke liye ulta chalaao.
Compton Effect — momentum book-keeping ( $p = h / λ$ ) ek collision mein le jaaya gaya.
Wave-Particle Duality — kyun Cell D ki huge photon count ek smooth wave jaisi dikhti hai.
Planck's Law and Blackbody Radiation — jahan $E = h f$ paida hua tha.