On a photosensitive material, when frequency of incident radiation is increased by 30% kinetic energy of emitted photo electrons increases from 0.4eV to 0.9eV. The work function of the surface is :
1.8eV
1.4eV
1eV
1.267eV
A
1eV
B
1.4eV
C
1.8eV
D
1.267eV
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Solution
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The correct option is B1.267eV
As per Einstein's Photoelectric equation:
K.E.max = hν−W
0.4 = h×ν−W --- (1)
0.9 = h×1.3ν−W --- (2)
(2) - (1)
0.5 = 0.3×hν⇒hν=1.667eV
Putting this value in equation (1),
W=1.667−0.4=1.267eV
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