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

WHAT: $E=h\nu$ apply karo. WHY: frequency di gayi hai, to seedha $\nu$ wala form use karo. $E=h\nu =(6.63\times 10^{-34})(6.0\times 10^{14})=3.98\times 10^{-19}\text{J}.$ eV mein convert karo $e$ se divide karke (kyunki $1\text{eV}=1.6\times 10^{-19}$ J): $E=\frac{3.98\times 10^{-19}}{1.6\times 10^{-19}}=2.49\text{eV}.$

WHAT: wavelength shortcut use karo. WHY: wavelength nm mein di gayi hai, to $hc=1240$ eV·nm ek hi step mein convert kar deta hai. $E=\frac{hc}{\lambda }=\frac{1240\text{eV⋅nm}}{620\text{nm}}=2.0\text{eV}.$

WHAT: $\varphi =h{\nu }_0$ ko rearrange karo. WHY: threshold par photon ki energy bilkul escape toll ke barabar hoti hai — kuch bacha nahin rehta. Pehle $\varphi$ ko joules mein convert karo: $\varphi =2.0\times 1.6\times 10^{-19}=3.2\times 10^{-19}$ J. ${\nu }_0=\frac{\varphi }{h}=\frac{3.2\times 10^{-19}}{6.63\times 10^{-34}}=4.83\times 10^{14}\text{Hz}.$

Step 1 — photon energy. $E=\frac{1240}{450}=2.756\text{eV}$. Step 2 — Einstein's equation. WHY: photon jo bhi laata hai minus toll — wahi bacha hua kinetic energy hai. $K{E}_{\max }=E-\varphi =2.756-2.3=0.456\text{eV}.$ Step 3 — stopping potential. Kyunki $e{V}_0=K{E}_{\max }$ hai, aur eV mein kaam karne par charge $e$ numerically cancel ho jaata hai: $V_0=0.456\text{V}.$

Step 1 — work function. $\varphi =\frac{hc}{{\lambda }_0}=\frac{1240}{300}=4.13\text{eV}.$ Step 2 — green light test karo. Uski photon energy hai $\frac{1240}{550}=2.25\text{eV}$. Compare karo: $2.25\text{eV}<4.13\text{eV}$, to ek green photon toll nahin chuka sakta. Chahe kitni bhi bright light ho, koi emission nahin. Rule yeh hai: zyaada wavelength = kam energy, aur $\lambda >{\lambda }_0$ hamesha fail karta hai.

Step 1 — photon energy. $E=\frac{1240}{200}=6.2\text{eV}$. Step 2 — Einstein's equation ko ulta karo. WHY: $e{V}_0=K{E}_{\max }$ hai, aur $K{E}_{\max }=E-\varphi$, to $\varphi =E-e{V}_0$. eV mein, $e{V}_0$ numerically sirf $2.2$ eV hai: $\varphi =6.2-2.2=4.0\text{eV}.$

Step 1 — slope. $V_0=\frac{h}{e}\nu -\frac{\varphi }{e}$ se, slope hai $h/e$. Figure mein red line dekho — uska rise over run hai: $\text{slope}=\frac{1.96-0.30}{(9.0-5.0)\times 10^{14}}=\frac{1.66}{4.0\times 10^{14}}=4.15\times 10^{-15}\text{V⋅s}.$ Step 2 — Planck's constant. $h=\text{slope}\times e=4.15\times 10^{-15}\times 1.6\times 10^{-19}=6.64\times 10^{-34}\text{J⋅s}.$ ✓ Step 3 — ek point se work function. Point 1 use karo: $V_{01}=\frac{h}{e}{\nu }_1-\frac{\varphi }{e}$, to $\frac{\varphi }{e}=\frac{h}{e}{\nu }_1-V_{01}=(4.15\times 10^{-15})(5.0\times 10^{14})-0.30=2.075-0.30=1.775\text{V}.$ Isliye $\varphi =1.775\text{eV}\approx 1.78\text{eV}$ ($\varphi /e$ volts mein $e$ se multiply karne par $\varphi$ eV mein numerically same rehta hai).

(a) Work functions. X-intercept threshold frequency ${\nu }_0=\varphi /h$ hai. B ki line zyaada right par cross karti hai ⇒ bada ${\nu }_0$ ⇒ bada $\varphi$ (B ke electrons ko free karna zyaada mushkil hai). (b) Slopes. Dono lines parallel hain aur slope $h/e$ hai — yeh ek universal ratio hai. To slopes identical hain; metal sirf intercepts badalta hai, tilt kabhi nahin. Yahi universality hai jisne Millikan ko $h$ nikalne diya, chahe koi bhi metal use kiya ho.

Step 1 — electron ki kinetic energy. $E_{\text{photon}}=\frac{1240}{300}=4.133\text{eV}$, to $K{E}_{\max }=4.133-2.0=2.133\text{eV}.$ Joules mein convert karo: $KE=2.133\times 1.6\times 10^{-19}=3.41\times 10^{-19}\text{J}$. Step 2 — KE se momentum. WHY: de Broglie ko momentum $p$ chahiye, aur slow (non-relativistic) electron ke liye $KE=\frac{p^2}{2m}$, to $p=\sqrt{2mKE}$. $p=\sqrt{2(9.1\times 10^{-31})(3.41\times 10^{-19})}=\sqrt{6.21\times 10^{-49}}=7.88\times 10^{-25}\text{kg⋅m/s}.$ Step 3 — de Broglie wavelength. ${\lambda }_{dB}=\frac{h}{p}$: ${\lambda }_{dB}=\frac{6.63\times 10^{-34}}{7.88\times 10^{-25}}=8.41\times 10^{-10}\text{m}=0.84\text{nm}.$

Step 1 — har photon test karo. $E_1=\frac{1240}{400}=3.10\text{eV}$; $E_2=\frac{1240}{250}=4.96\text{eV}$. Dono $\varphi =3.0$ eV se zyaada hain, to dono electrons eject karte hain (har photon independently kaam karta hai — ek photon, ek electron). Step 2 — sabse fast electron zyaada energetic photon se aata hai. WHY: $K{E}_{\max }$ photon energy ke saath badhta hai, to ${\lambda }_2$ (neela, chota) sabse fast electron deta hai. $K{E}_{\max }=E_2-\varphi =4.96-3.0=1.96\text{eV}.$ 400 nm light bhi electrons contribute karti hai, lekin slow wale ($3.10-3.0=0.10$ eV). Poore setup ka stopping potential sabse fast wale se set hota hai: $V_0=1.96$ V.

Step 1 — dono equations likho. WHY: do unknowns ($h\nu$ scale aur $\varphi$) ke liye do equations chahiye. $e{V}_0=h\nu -\varphi (1)$ $e{V}_0^{\prime }=h(2\nu )-\varphi (2)$ Step 2 — (1) ko (2) se subtract karo. $\varphi$ cancel ho jaata hai, $h\nu$ isolate ho jaata hai: $e(V_0^{\prime }-V_0)=h\nu ⟹h\nu =e(2.3-0.5)=1.8\text{eV}.$ (eV mein, $e(V_0^{\prime }-V_0)$ numerically $1.8$ eV hai.) Step 3 — (1) mein back-substitute karo. $\varphi =h\nu -e{V}_0=1.8-0.5=1.3\text{eV}.$ Step 4 — threshold frequency. ${\nu }_0=\frac{\varphi }{h}=\frac{1.3}{h}$. Kyunki $h\nu =1.8$ eV hai, to $h=1.8/\nu$ (eV per Hz-unit), isliye ${\nu }_0=\frac{1.3}{1.8/\nu }=\frac{1.3}{1.8}\nu =0.722\nu .$ To threshold original frequency ke lagbhag $72\%$ par hai — original light pehle se hi threshold se upar thi, jaise ki hona chahiye tha positive $V_0$ ke liye.

Claim galat hai. $\nu ={\nu }_0$ par: $K{E}_{\max }=h{\nu }_0-\varphi =\varphi -\varphi =0.$ Photon ki poori energy escape toll par kharch ho jaati hai; motion ke liye kuch nahin bachta. Electron bas barely surface tak zero speed se pahunchta hai. Limiting behaviour: jaise jaise $\nu \to {\nu }_0^{+}$ (threshold se upar se approach karte hue), $K{E}_{\max }=h(\nu -{\nu }_0)\to 0^{+}$ — electrons dhire aur dhire nikalte hain, aur stopping potential $V_0\to 0$. ${\nu }_0$ se neeche, $K{E}_{\max }$ negative hoga, jo physically impossible hai — matlab koi electron escape nahin karta (equation apply karna band ho jaati hai; emission simply switch off ho jaata hai).

Step 1 — surface par power. $A=1.0{\text{cm}}^2=1.0\times 10^{-4}{\text{m}}^2$. $P=IA=(1.0\times 10^{-3})(1.0\times 10^{-4})=1.0\times 10^{-7}\text{W}.$ Step 2 — har photon ki energy (joules mein). $E=\frac{hc}{\lambda }=\frac{(6.63\times 10^{-34})(3.0\times 10^8)}{400\times 10^{-9}}=4.97\times 10^{-19}\text{J}.$ Step 3 — photons per second. WHY: power energy/second hai, to energy-per-photon se divide karne par photons/second milte hain. $N=\frac{P}{E}=\frac{1.0\times 10^{-7}}{4.97\times 10^{-19}}=2.01\times 10^{11}\text{photons/s}.$ Step 4 — current. Har photon se ek electron, har electron charge $e$ carry karta hai: $I_{\text{current}}=Ne=(2.01\times 10^{11})(1.6\times 10^{-19})=3.22\times 10^{-8}\text{A}\approx 32\text{nA}.$ Note: $\varphi =2.0$ eV ek distractor tha — yeh sirf decide karta hai ki emission hogi ya nahin (yahan $E=3.10$ eV $>\varphi$, to haan), current magnitude nahin jab har photon succeed kare.

$V_0$ vs $\nu$ graph ka universal slope ==$h/e$== hai. Bilkul threshold frequency par, $K{E}_{\max }$ zero ke barabar hota hai. Do wavelengths mein sabse fast electron dhundhne ke liye, shorter (more energetic) wavelength use karo. Ejected electron ki de Broglie wavelength ${\lambda }_{dB}=$ ==$h/p$== hai jahan $p=\sqrt{2mKE}$.