First slide

Photo Electric Effect

Question

A photon of energy ‘2’ ev is incident on a metal, The minimum de-Broglie wave length of the emitted electrons is $λ$ . When the wave length of the incident radiation is reduced by 60% the minimum de-Broglie wave length of the emitted electrons is $\frac{λ}{2}$ The work function of the metal is

Difficult

A

1.0 ev
B

0.5 ev
C

0.75 ev
D

1.5 ev

Solution

if de Broglie wavelength becomes halved momentum becomes doubled and kinetic energy becomes 4 times and if incident wavelength decreases by 60% it means it becomes 40%
$λ b e c o m e s \frac{4}{10} λ f r e q u e n c y b e c o m e s \frac{10}{4} ν E i e n s t e i n e q u a t i o n h ν = w + k - - - - (1) s e c o n d t i m e h \frac{10}{4} ν = w + 4 k - - - - (2) s o l v i n g w = 0.5 h ν h e r e h ν = 2 e v$

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