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)

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

Answer 1

Answer: (C)

$\quad\frac{hc}{2 \lambda}$

Answer 2

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 $K_{max2}= 3K_{max1} \, (given)$
$\therefore \, \, \, \frac {2hc}{? }-\phi_0=3 \bigg ( \frac {hc}{?}-\phi_0 \bigg )$
$\frac {2hc}{?}-\phi_0=\frac {3hc}{?}-3 \phi_0$
$3\phi_0-\phi_0= \frac {3hc}{?}- \frac {2hc}{?}$
$2\phi _0=\frac {hc}{?} \, \, \, \, or \, \phi_0= \frac {hc}{2?}$