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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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Light described at a place by the equation $\;\varepsilon=(100 V/m)\;[sin(5\times10^{15} s^{-1})t+sin(8 \times10^{15})t]\;$ falls on a metal surface having work function 2eV . Than maximum kinetic energy of photo electrons will be :

$(a)\;5.2 eV\qquad(b)\;4.17 eV\qquad(c)\;3.27 eV\qquad(d)\;6.2 eV$

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Answer : (c) 3.27 eV
Explanation :
The light contains two different frequencies . The one with larger frequency will cause photo electrons with largest kinetic energy .
This larger frequency is
$\nu=\large\frac{w}{2 \pi}=\large\frac{8\times10^{15}}{2 \pi}\;s^{-1}$
The maximum kinetic energy of photo electrons is $\;K_{max}=h \nu-W$
$=(4.14\times10^{-15}eV -s)\times(\large\frac{8\times10^{15}}{2 \pi}s^{-1})-2eV$
$=5.27eV-2eV=3.27eV\;.$
answered Feb 22, 2014 by yamini.v
 

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