> Q. Ultraviolet light of wavelength 200 nm is incident on polished surface of Fe (iron). Work function of the surface is 4.71 eV. What will be its stopping potential? $ ( h = 6.626 times 10^{ – 34} , J-s; , 1 e , V = 1.6 times 10^{ – 19 } , J$, c = $ 3 times 10^8 , ms^{ – 1} ) $ – LIVE ANSWER TODAY

Q. Ultraviolet light of wavelength 200 nm is incident on polished surface of Fe (iron). Work function of the surface is 4.71 eV. What will be its stopping potential? $ ( h = 6.626 times 10^{ – 34} , J-s; , 1 e , V = 1.6 times 10^{ – 19 } , J$, c = $ 3 times 10^8 , ms^{ – 1} ) $

In This post you will get answer of Q.
Ultraviolet light of wavelength 200 nm is incident on polished surface of Fe (iron). Work function of the surface is 4.71 eV. What will be its stopping potential?
$ ( h = 6.626 times 10^{ – 34} , J-s; , 1 e , V = 1.6 times 10^{ – 19 } , J$,
c = $ 3 times 10^8 , ms^{ – 1} ) $

Solution:

Given that, the wavelength of incident light

$ ( lambda ) = 200 ,nm $

= $ 200 times 10^{ – 9 } , m $

work function ( $ phi $) = 4.71 eV

From Einstein’s photoelectric equation

$ KE_{ max} = hv – phi $ …(i)

But $ KE_{ max } = e V_0 $ …(ii)

So, $ e V_0 = hv – phi $ $(V_0 = $ Stopping potential)

or $ e V_0 = frac{ hc }{ lambda } – phi $

$ therefore 1.6 times 10^{ – 19 } V_0 = frac{ 6.6 times 10^{ – 34} times 3 times 10^8 – 4.71 times 1.6 times 10^{ – 19 }}{ 200 times 10^{ – 9 }} $

$ Rightarrow V_0 = frac{ 6.6 times 10^{ – 34} times 3 times 10^8 }{ 2 times 10^{ – 7 } times 1.6 times 10^{ – 19 }} – frac{ 4.71 times 1.6 times 10^{ – 19 }}{ 1.6 times 10^{ – 19 }} $

$ Rightarrow V_0 = ( 6.19 – 4.71 ) $

$ Rightarrow V_0 = 1.48 $

$ Rightarrow V_0 approx 1.50 $ V

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