Question:

A photoelectric surface is illuminated successively by monochromatic light of wavelength $? $ and 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 of the material is (h = Planck's constant, c = speed of light)

Updated On: Jun 23, 2024
  • $\frac {2hc}{?} $
  • $\frac {hc}{3?} $
  • $\quad\frac{hc}{2 \lambda}$
  • $\frac {hc}{?} $
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The Correct Option is C

Solution and Explanation

Let $\phi_0 $ be the work function of the surface of the material. Then,
According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is
$K_{max_1}= \frac {hc}{?}-\phi_0$
and that in the second ease is
$K_{max_2}= \frac {hc}{?}-\phi_0= \frac {2hc}{?}-\phi_0 $
But $K_{max2}= 3K_{max1} \, (given) $
$\therefore \, \, \, \frac {2hc}{? }-\phi_0=3 \bigg ( \frac {hc}{?}-\phi_0 \bigg )$
$\frac {2hc}{?}-\phi_0=\frac {3hc}{?}-3 \phi_0$
$3\phi_0-\phi_0= \frac {3hc}{?}- \frac {2hc}{?}$
$2\phi _0=\frac {hc}{?} \, \, \, \, or \, \phi_0= \frac {hc}{2?} $
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