Question

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is 4.2 eV. In order to reduce the photo-current to zero the voltage of the anode relative to the cathode must be made

• A. – 4.2 V
• B. – 9.4 V
• C. – 17.8 V
• D. +9.4 V

Answer 1

The correct option is B – 9.4 V

$E=W_0+e{V}_0$
For hydrogen atom, E=+13.6eV
$\therefore$ $+13.6=4.2+e{V}_0$
$\Rightarrow V_0=\frac{(13.6-4.2)eV}{e}=9.4V$
Potential at anode = - 9.4 V