Question

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} ) $

• A. 0.5 V
• B. 2.5 V
• C. 1.5 V
• D. None of these

Answer 1

Answer: (C)

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