Question

Light from a hydrogen discharge tube is incident on the cathode of a photocell. 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:

• A. $-8.4V$
• B. $-4.2V$
• C. $-2.1V$
• D. $9.4V$

Answer 1

The correct option is D $9.4V$

Given that,

Work function = $4.2eV$

Energy of photon from $H$ atom is given by the difference between the two energy levels in which electron transit

We know that, the energy at $n^{th}$ level is

$E_n=\frac{-13.6}{n^2}$

Now, maximum energy will be

$E=\frac{13.6}{n_i^2}-\frac{13.6}{n_f^2}$

$E=\frac{13.6}{1}-0$

$E=13.6eV$

Now, using Einstein equation

Maximum K.E = energy of the incident radiation –work function

$K.E=13.6-4.2$

$K.E=9.4eV$

Hence, the potential of the anode is $9.4V$