Photon properties — E = hf, p = h - λ: Question bank

2.3.3 · D5 · Physics › Modern Physics › Photon properties — E = hf, p = h - λ

Shuru karne se pehle, neeche use hone wale saare symbols ek jagah, taaki kuch assumed na ho:

True or false — justify karo

Ek photon ki frequency double karne se uski energy double hoti hai.

True.

E = h f

seedha proportionality hai —

E

aur

f

bilkul ek saath badhte hain, toh double frequency matlab exactly double energy.

Ek photon ki wavelength double karne se uska momentum double hota hai.

False.

p = h / λ

matlab

p \propto 1/ λ

;

λ

double karne se

p

aadha ho jaata hai. Momentum lambi (lal) waves ke liye ghatta hai, badhta nahi.

Usi colour ki zyada bright beam mein zyada energetic photons hote hain.

False. Same colour = same

f

= same energy per photon. "Brighter" ka matlab sirf zyada photons per second hai; har ek abhi bhi exactly

h f

hi carry karta hai.

Same wavelength ke do photons hamesha same momentum carry karte hain.

True. Momentum sirf

λ

par depend karta hai

p = h / λ

ke through; identical wavelength forced identical momentum deti hai, chahe source ya brightness koi bhi ho.

Vacuum mein ek photon ko less energetic banake

c

se neeche slow kiya ja sakta hai.

False. Vacuum mein har photon exactly

c

par hi move karta hai chahe energy kuch bhi ho. Energy kam karna

f

kam karta hai (lal light), speed nahi — speed relativity se fixed hai.

Kyunki photon massless hai, uska momentum zero hai.

False.

p = m v

sirf low-speed limit hai. Asli relation

E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}

mein

m_{0} = 0

rakhne par

p = E / c \neq = 0

milta hai. Massless ≠ momentumless.

E = h f

aur

E = h c / λ

ek photon ke baare mein do alag independent facts hain.

False. Ye same equation hai,

c = f λ

se connected.

f = c / λ

substitute karne par ek doosre mein convert ho jaati hai; tum dono ko ek saath kabhi use nahi karte.

Ek photon ke liye,

E = p c

exactly hold karta hai, bina kisi correction term ke.

True. Rest-mass term

(m_{0} c^{2})^{2}

zero ho jaata hai kyunki

m_{0} = 0

, sirf

E = p c

bachta hai bina kisi leftover ke, unlike massive particle jisme woh term rehta hai.

eV aur joules mein ek photon ki energy do alag physical quantities describe karti hain.

False. Ye same energy hai do units mein; tum joules ko

1.6 \times 1 0^{- 19}

se divide karke convert karte ho. eV sirf chhoti energies ke liye ek human-friendly ruler hai.

Error pakdo

p = m v

, aur photon ke liye

m = 0

, toh photon solar sail par koi force nahi lagaata."

Error yeh hai ki

p = m v

ko uski valid low-speed range ke bahar use kar rahe hain. Sahi

p = h / λ

nonzero hai, isliye solar sails aur comet tails ko light ka push feel hota hai (radiation pressure).

"Photons per second nikaalane ke liye main power ko energy per photon se multiply karta hoon:

N = P \times (h c / λ)

Error multiply karne ki jagah divide karna hai. Power = (energy per photon) × (photons per second), toh

N = P \div (h c / λ)

. Multiply karne par W·J ki units aati hain, jo yahaan meaningless hai.

"500 nm photon ka

E = h f = h \times 500 \times 1 0^{- 9}

joules hai."

Error yeh hai ki frequency slot mein wavelength daali ja rahi hai.

f

Hz mein hota hai, metres mein nahi. Ya pehle

f = c / λ

use karo, ya seedha

E = h c / λ

form switch karo.

"Red light photons aur blue light photons mein same energy hoti hai kyunki dono sirf light hain."

Error yeh hai ki energy frequency ke saath scale karti hai, yeh ignore kiya gaya. Blue ka

f

red se zyada hai, toh har blue photon zyada energy carry karta hai — "sirf light" ek real per-photon difference chhupata hai.

"Kyunki

E = h f

hai aur momentum ke liye mass chahiye, toh momentum

E

times mass hona chahiye."

Do errors hain: momentum ke liye mass ki zaroorat nahi, aur invented formula dimensionally galat hai. Sahi link hai

p = E / c

, momentum equals energy divided by speed of light.

"Agar main same wavelength par laser ko twice as powerful banata hoon, toh har photon double energetic ho jaata hai."

Error total power aur per-photon energy ko confuse karna hai. Same

λ

har photon ki energy fix kar deta hai; double power matlab double photons per second, mote photons nahi.

Why questions

Ek single photon ki energy frequency par depend kyun karti hai, amplitude par nahi?

Ek akele photon ka koi meaningful amplitude nahi hota — tum "aadha photon" ya "louder" photon nahi rakh sakte. Frequency hi ek red lump aur ek blue lump ko distinguish karne ka knob hai, isliye energy

f

par sawaar rehni chahiye; amplitude sirf photon count ke roop mein bachta hai.

Photon ke liye hum

p = m v

ki jagah relativistic relation

E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}

kyun use karte hain?

Kyunki

p = m v

ek non-relativistic approximation hai aur

m_{0} = 0

hone par fail kar deta hai, galat answer "zero" deta hai. Relativistic relation hi general truth hai aur massless particle ke liye

c

par move karne par sahi

p = E / c

deta hai.

Photoelectric effect light ki frequency ki parwah karta hai, brightness ki nahi — kyun?

Electron nikaalane ke liye ek single photon ki kaafi energy

h f

chahiye jo metal ki binding ko beat kar sake. Brightness zyada photons add karti hai par har ek mein zyada energy nahi, toh threshold frequency se neeche kitni bhi brightness kaam nahi karti.

Hum energy formulas ke beech

f = c / λ

itni freely substitute kyun kar sakte hain?

Kyunki vacuum mein saari light wave relation

c = f λ

follow karti hai,

f

aur

λ

ko ek fixed reciprocal pair mein lock karti hai. Ek jaanna doosra deta hai, isliye dono energy forms interchangeable hain, kabhi independent nahi.

Light kisi surface par dabaav kyun daalti hai jab woh usse hit karti hai?

Har photon momentum

p = h / λ

carry karta hai; use absorb ya reflect karna woh momentum surface ko transfer karta hai. Photons ki ek stream jo per second bahut saari chhoti pushes deliver karti hai, steady force mein add ho jaati hai — radiation pressure.

Planck's constant

h

dono

E

aur

p

ke liye itna central kyun hai?

h

ek wave ki frequency/wavelength aur uski particle-like energy/momentum ke beech universal conversion rate hai. Yeh single "exchange rate" hai jo wave language ko dono formulas mein quantum-lump language mein convert karta hai.

Edge cases

Kisi photon ka momentum kya hoga jab uski wavelength bina bound ke badhti jaaye (bahut low frequency radio)?

Jab

λ \to \infty

p = h / λ \to 0

aur

E = h c / λ \to 0

. Aisa photon vanishingly thodi energy aur push carry karta hai, halanki woh abhi bhi exist karta hai aur abhi bhi

c

par move karta hai.

Kya ek photon ki energy exactly zero ho sakti hai?

Nahi. Uske liye

f = 0

λ = \infty

chahiye hoga, matlab bilkul oscillation nahi — photon nahi. Har real photon ka

f > 0

hota hai, isliye strictly positive

E

aur

p

hota hai.

Agar tum ek photon ko ek door jaate frame se dekhte ho (redshift), toh uski energy aur momentum ka kya hoga?

Dono saath mein ghatte hain. Frequency neeche shift hoti hai, toh

E = h f

ghatta hai, aur kyunki

p = E / c

hai momentum exactly usi proportion mein ghatta hai — energy aur momentum har frame mein

E = p c

se locked rehte hain.

Kya vacuum mein

c

ke alawa koi photon speed hai?

Nahi. Massless particles vacuum mein sabhi inertial frames mein exactly

c

par travel karne ke liye forced hain; photon ka koi rest frame nahi hota, isliye "stationary photon" ek contradiction hai.

Agar do photons combine hote hain (jaise absorption mein), kya unki energies simply add hoti hain?

Unki energies aur momenta conservation quantities ke roop mein add hote hain, lekin result generally ek aur single photon nahi hota — energy aur momentum conservation milke ek massless lump ko combined amounts carry karne se rokti hain jab tak woh identically travel na karein. Bookkeeping ke liye Compton Effect aur Relativistic Energy-Momentum Relation dekho.

Photon ke liye

p = h / λ

"forward" run karna aur electron ke liye "backward" karna — dono mein kya farq hai?

Photon ke liye hum light ki

λ

se shuru karke uska momentum nikalte hain. Matter ke liye, de Broglie Wavelength ise ulta chalata hai — diye gaye particle ke momentum

p

se, use

λ = h / p

ki wave assign karo. Same constant, opposite direction.

Recall Page band karne se pehle ek-line self-test

Sab kuch cover karo: batao kyun brightness photon count change karti hai na ki photon energy, aur kyun ek massless photon phir bhi ek wall ko dhakka deta hai. Answer ::: Har photon ki energy frequency se fix hoti hai ( $E = h f$ ), toh zyada brightness matlab sirf zyada photons; aur momentum $p = E / c$ se aata hai ( $p = m v$ se nahi), jo $m_{0} = 0$ hone par bhi nonzero rehta hai, isliye har photon ek real push deliver karta hai.

Connections

Photoelectric Effect — frequency-not-brightness trap experimentally proven.
Compton Effect — "massless means no momentum" trap seedha disprove hua.
Relativistic Energy-Momentum Relation — $E = p c$ reasoning ka ghar.
de Broglie Wavelength — matter ke liye $p = h / λ$ ulta chalaaya gaya.
Radiation Pressure — light ke surfaces ko push karne ka edge case.
Wave-Particle Duality — woh badi picture jiske andar ye traps rehte hain.