Question

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is $4.2eV$. In order to reduced the photocurrent to zero, the voltage of the anode relative to the cathode must be made

• A. $-4.2V$
• B. $-9.4V$
• C. $-17.8V$
• D. $+9.4V$

Answer 1

The correct option is B $-9.4V$

Photoelectric Law:
$E=W_o+e{V}_o$

For Hydrogen atom:

$E=-13.6\left(\frac{1}{n^2}\right)eV$

For $n=1$, $E=13.6eV$
$13.6=4.2+e{V}_o$
$V_o=-9.4V$

Potential at anode $=-9.4V$