Question

the photo electrons emitted by light of ware length 4000A can be stopped by a retarding potential of 2volt find the topping potential for light of warelength 3000A​

Answer 1

$\sf \bold{Given\::}$

$\sf \bullet$ When wavelength = 4000Å , stopping potential = 2V

$\sf \bold{To\:Find\::}$ Stopping potential when wavelength = 3000Å

$\sf \bold{Formula\:used\::}$ $\sf eV\:=\:\dfrac{hc}{x} - W$

Here,

$\sf \bullet$ e = Charge of electron

$\sf \bullet$ V = stopping potential

$\sf \bullet$ h = Planck's constant

$\sf \bullet$ c = speed of light in vaccum

$\sf \bullet$ x = wavelength

$\sf \bullet$ W = work function

$\sf \bold{Solution\::}\\When\:wavelength\: =\: 4000Å ,\:stopping\:potential\:=\:2V\\\\\sf \implies eV = \dfrac{hc}{x} - W\\\\ \sf \implies W = \dfrac{hc}{x} - eV\\\\\sf \implies W = \dfrac{6.6\times 10^{-34}\: \times 3\:\times 10^{8}\:}{4000\times\:10^{-10}}\:-\: ( 1.6\:\times\:10^{-19}\:\times\:2)\\\\\\\sf \implies W = 4.95 \times \: 10^{-19} \: \:- \:3.2 \times 10^{-19}\\\\\sf \implies W = (4.95 \: \:- \:3.2 )\times 10^{-19}\\\\\sf \implies W = 1.75\: \times 10^{-19}$

Work function remain same for metal.

When wavelength is 3000Å = 3000×$10^{-10}$m

$\sf \implies eV' = \:\dfrac{hc}{x} - W\\\\\sf \implies V' =\: (\dfrac{1}{e} )\times(\dfrac{hc}{x} - W)\\\\\\\sf \implies V' =\: (\dfrac{1}{e} )\times(\dfrac{6.6\times 10^{-34}\: \times 3\:\times 10^{8}\:}{3000\:\times\:10^{-10}} -\:1.75\:\times 10^{-19} )\\\\\\\sf \implies V' =\: (\dfrac{1}{e} )\times(\:6.6\:\times\:10^{-19}\:-\:1.75\:\times 10^{-19} )\\\\\\\sf \implies V' =\: (\dfrac{1}{1.6\:\times\:10^{-19}} )\times(\:6.6\:-\:1.75)\times 10^{-19}$

$\sf \implies V' =\: (\dfrac{10^{-19} }{1.6\:\times\:10^{-19}} )\times(\:4.85\:)\\\\\\\sf \implies V' =\: (\dfrac{4.85 }{1.6} ) \\\\\sf \implies \bold{ V'\:=\: 3.03volt }$

$\sf \bold{ ANSWER\::} \: Stopping\: Potential\:=\: \bold{3.03volt}$