JEE Main 12 Jan 2019 Morning Question 2
Question: What is the work function of the metal if the light of wavelength $ 4000\overset{o}{\mathop{A}}, $ generates photoelectrons of velocity $ 6\times 10^{5}m,{s^{-1}} $ from it?
(Mass of electron $ =9\times {10^{-31}}kg, $ Velocity of light $ =3\times 10^{8}m,{s^{-1}}, $ Planck?s constant $ =6.626\times {10^{-34}}Js, $ Charge of electron $ =1.6\times {10^{-19}}Je{V^{-1}} $ ) [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) 4.0 eV
B) 2.1 Ev
C) 0.9 eV
D) 3.1 eV
Show Answer
Answer:
Correct Answer: B
Solution:
Given, $ \lambda =4000\overset{o}{\mathop{A}}, $ $ v=6\times 10^{5},m,{s^{-1}} $
$ h\upsilon =w _0+\frac{1}{2}mv^{2} $
[ $ w _0= $ work function] $ w _0=h\upsilon -\frac{1}{2}mv^{2} $
$ =\frac{6.626\times {10^{-34}}\times 3\times 10^{8}}{4000\times {10^{-10}}}-\frac{1}{2}\times 9\times {10^{-31}}\times {{(6\times 10^{5})}^{2}} $
$ =(4.9695-1.62){10^{-19}}=3.3495\times {10^{-19}}J $ $ w _0=\frac{3.3495\times {10^{-19}}}{1.6\times {10^{-19}}}eV=2.0934\approx 2.1eV $
