Dual Nature Of Light Question 408
Question: A photoelectric surface is illuminated successively by monochromatic light of wavelengths $ \lambda $ and $ \frac{\lambda }{2}. $ If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is:
Options:
A) $ \frac{hc}{2\lambda } $
B) $ \frac{hc}{\lambda } $
C) $ \frac{hc}{3\lambda } $
D) $ \frac{3hc}{\lambda } $
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Answer:
Correct Answer: A
Solution:
[a] From Einstein’s photoelectric equation $ K.E{._{\lambda }}=\frac{hc}{\lambda }-\phi $ ?.(i)
(for monochromatic light of wavelength $ \lambda $ ) where $ \phi $ is work function
$ K.E{._{\lambda /2}}=\frac{hc}{\lambda /2}-\phi $ …(ii)
(for monochromatic light of wavelength $ \lambda /2 $ )
From question, $ K.E._{\lambda /2}$
=$3 \times K.E._\lambda $
$\Rightarrow \frac{hc}{\lambda /2}-\phi =3( \frac{hc}{\lambda }-\phi ) $
$ \frac{2hc}{\lambda }-\phi =3\frac{hc}{\lambda }-3\phi \Rightarrow 2\phi =\frac{hc}{\lambda }\therefore \phi =\frac{hc}{2\lambda } $
BETA
