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

WHAT hum chahte hain: frequency se energy → $E=hf$ use karo (woh version jo seedha $f$ leta hai). $E=hf=(6.626\times 10^{-34})(6.0\times 10^{14})=3.98\times 10^{-19}\text{J}.$ WHY yeh form: humein $f$ diya gaya tha, isliye $\lambda$ ke through detour lene ki zaroorat nahi.

WHAT hum chahte hain: wavelength se momentum → sabse short route hai $p=h/\lambda$. Pehle convert karo: $400\text{nm}=400\times 10^{-9}\text{m}=4.0\times 10^{-7}\text{m}$. $p=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{4.0\times 10^{-7}}=1.66\times 10^{-27}\text{kg⋅m/s}.$

Step 1 — WHAT: $\lambda$ diya hai, energy chahiye → $E=hc/\lambda$ use karo. $E=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3\times 10^8)}{0.10\times 10^{-9}}=\frac{1.988\times 10^{-25}}{1.0\times 10^{-10}}=1.988\times 10^{-15}\text{J}.$ Step 2 — WHY convert: itne chhote joules padhne laayak nahi hain; eV paane ke liye $1.6\times 10^{-19}$ se divide karo. $E=\frac{1.988\times 10^{-15}}{1.6\times 10^{-19}}\approx 1.24\times 10^4\text{eV}=12.4\text{keV}.$ Chhoti wavelength ⇒ high energy — exactly isliye X-rays flesh mein ghus jaati hain.

WHAT: humein $p$ diya hai aur $\lambda$ chahiye → $p=h/\lambda$ ko $\lambda =h/p$ mein invert karo. $\lambda =\frac{h}{p}=\frac{6.626\times 10^{-34}}{2.2\times 10^{-27}}=3.01\times 10^{-7}\text{m}=301\text{nm}.$ Kya yeh visible hai? Visible band roughly $400$–$700\text{nm}$ hoti hai. $301\text{nm}$ chhota hai, isliye yeh ultraviolet hai, humari aankhon se thoda bahar.

Step 1 — energy per photon: $E=hc/\lambda =\frac{1.988\times 10^{-25}}{532\times 10^{-9}}=3.74\times 10^{-19}\text{J}.$ Step 2 — WHY divide: power = (energy per photon)$\times$(photons per second), isliye $N=P/E$. $N=\frac{P}{E}=\frac{2.0\times 10^{-3}}{3.74\times 10^{-19}}\approx 5.35\times 10^{15}\text{photons/s}.$

Reasoning: dono $E$ aur $p$ $1/\lambda$ ke anusaar chalte hain. Isliye ratio bas wavelengths ka ulta ratio hai. $\frac{E_B}{E_A}=\frac{{\lambda }_A}{{\lambda }_B}=\frac{600}{200}=3,\frac{p_B}{p_A}=\frac{600}{200}=3.$ Photon B (chhota $\lambda$) mein energy 3× aur momentum bhi 3× hai. Dono ko ek hi tarah scale karna $E=pc$ ki pehchaan hai.

WHAT: do photons ki energies ka ratio → $E\propto 1/\lambda$, isliye factor $={\lambda }_{\text{radio}}/{\lambda }_{\text{gamma}}$. $\frac{E_{\gamma }}{E_{\text{radio}}}=\frac{3.0}{1.0\times 10^{-12}}=3.0\times 10^{12}.$ Gamma photon teen trillion guna zyada energetic hai — wahi equation, spectrum ke opposite ends.

WHY yeh tool: photon apni saari energy $E$ ek lump mein deta hai; electron $\varphi$ escape karne mein kharach karta hai aur baaqi kinetic energy ke roop mein rakhta hai. Yahi Einstein ki photoelectric equation hai $K_{\max }=E-\varphi$ — Photoelectric Effect dekho. $K_{\max }=E-\varphi =2.48-2.0=0.48\text{eV}.$ Agar $E<\varphi$, koi bhi electron escape nahi karta — lump simply kaafi bada nahi hota, chahe kitne bhi zyada lumps aayein.

Algebra (WHY): parent ke do definitions se shuru karo aur $\lambda$ eliminate karo. $E=\frac{hc}{\lambda },p=\frac{h}{\lambda }⟹\frac{E}{p}=\frac{hc/\lambda }{h/\lambda }=c⟹E=pc.$ Yahi result Relativistic Energy-Momentum Relation bhi deta hai jab tum rest mass $m_0=0$ set karte ho. Numeric check ($\lambda =500\text{nm}$): $E=\frac{1.988\times 10^{-25}}{5.0\times 10^{-7}}=3.98\times 10^{-19}\text{J}$, aur $p=\frac{6.626\times 10^{-34}}{5.0\times 10^{-7}}=1.33\times 10^{-27}\text{kg⋅m/s}$. Phir $pc=(1.33\times 10^{-27})(3\times 10^8)=3.98\times 10^{-19}\text{J}=E.$ ✓

WHY momentum enter karta hai: force = rate of momentum delivered. Har photon $p=E/c$ carry karta hai. Ek second mein beam energy $P\times 1\text{s}$ deliver karti hai, isliye per second momentum $P/c$ hota hai. $F=\frac{\Delta p}{\Delta t}=\frac{P}{c}=\frac{1.0}{3\times 10^8}=3.33\times 10^{-9}\text{N}.$ Bahut chhota, lekin real hai — yahi Radiation Pressure hai, woh force jo solar sails ko push karta hai. (Ek reflecting panel ke liye yeh $2P/c$ ho jaata kyunki momentum reverse hota hai.)

Key insight: de Broglie Wavelength relation $p=h/\lambda$ electron ke liye bhi wahi hai jo photon ke liye hai — de Broglie ne photon rule ko matter ke liye ulta chalaya. $p_{\text{photon}}=\frac{h}{\lambda }=\frac{6.626\times 10^{-34}}{1.0\times 10^{-10}}=6.63\times 10^{-24}\text{kg⋅m/s}.$ Kyunki dono ek hi $\lambda$ share karte hain, dono ek hi momentum share karte hain: $p_{\text{electron}}=6.63\times 10^{-24}\text{kg⋅m/s}$ bhi. Unki energies bahut alag hain (electron mein rest mass hai, photon mein nahi), lekin $p$ sirf $\lambda$ se fixed hota hai.

(a) Photons per second. Energy per photon: $E=hc/\lambda =\frac{1.988\times 10^{-25}}{550\times 10^{-9}}=3.61\times 10^{-19}\text{J}.$ $N=\frac{P}{E}=\frac{100}{3.61\times 10^{-19}}\approx 2.77\times 10^{20}\text{photons/s}.$ (b) Mirror par force. Mirror tak pahunchne wali power: $P^{\prime }=0.010\times 100=1.0\text{W}$. Reflection ⇒ momentum reverse hota hai ⇒ force $=2P^{\prime }/c$. $F=\frac{2P^{\prime }}{c}=\frac{2(1.0)}{3\times 10^8}=6.67\times 10^{-9}\text{N}.$ $2.0\text{m}$ distance ek decoy hai — yeh enter nahi karta, kyunki humein directly bataya gaya tha ki kitni fraction light pakdi jaati hai.

Yeh Compton Effect hai: ek photon ek solid ball ki tarah behave karta hai jo recoiling electron ko energy deta hai. (a) Photon se lose hua energy = electron ka gain hua energy. Dono photon energies $E=hc/\lambda$ se compute karo. $E=\frac{1.988\times 10^{-25}}{0.0700\times 10^{-9}}=2.840\times 10^{-15}\text{J},$ $E^{\prime }=\frac{1.988\times 10^{-25}}{0.0724\times 10^{-9}}=2.746\times 10^{-15}\text{J}.$ Electron ka gain $=E-E^{\prime }=2.840\times 10^{-15}-2.746\times 10^{-15}=9.42\times 10^{-17}\text{J}\approx 589\text{eV}.$ (b) Proof: photon ka wavelength badal gaya (isliye momentum $p=h/\lambda$ bhi). Ek pure wave bounce karke apna wavelength waisa hi rakhti. Wavelength shift ka matlab hai photon ne momentum carry kiya aur ek particle ki tarah trade kiya — momentum smoking gun hai.

Mujhe $\lambda$ do, main $E$ kaise nikaaloon? ::: $E=hc/\lambda$. Mujhe $\lambda$ do, main $p$ kaise nikaaloon? ::: $p=h/\lambda$. Power $P$ ki absorbed beam ki force? ::: $F=P/c$. Power $P$ ki reflected beam ki force? ::: $F=2P/c$. Power $P$, wavelength $\lambda$ se photons per second? ::: $N=P/(hc/\lambda )$. Chhoti wavelength ka matlab ___ energy aur ___ momentum. ::: zyada; zyada (dono $\propto 1/\lambda$).