Question

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ 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)

• A. $\frac {2hc}{\lambda} $
• B. $\frac {hc}{3\lambda} $
• C. $\frac{hc}{2 \lambda}$
• D. $\frac {hc}{\lambda} $

Answer 1

Answer: (C)

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
$K_{\max _{1}}=\frac{h c}{\lambda}-\phi_{0}$
and that in the second ease is
$K_{\max _{2}}=\frac{h c}{\lambda}-\phi_{0}=\frac{2 h c}{\lambda}-\phi_{0}$
But $K_{\max 2}=3 K_{\max 1}($ given $)$
$\therefore \frac{2 h c}{\lambda}-\phi_{0}=3\left(\frac{h c}{\lambda}-\phi_{0}\right) $
$\frac{2 h c}{\lambda}-\phi_{0}=\frac{3 h c}{\lambda}-3 \phi_{0} $
$3 \phi_{0}-\phi_{0}=\frac{3 h c}{\lambda}-\frac{2 h c}{\lambda} $
$2 \phi_{0}=\frac{h c}{\lambda} $
or $ \phi_{0}=\frac{h c}{2 \lambda}$