Question

Light from a hydrogen discharge tube is incident on the cathode of a photo-cell. The work function of the cathode surface is 3.8 eV. In order to reduce the photo-electric current to zero, the voltage of anode relative to cathode must be

• A. +9.8V
• B. -9.8 V
• C. +0.4 V
• D. -2.3 V

Answer 1

The correct option is B -9.8 V

The maximum kinetic energy of the ejected electron is :
$K_{max}=hf-\varphi$, where, $\varphi$ is the work function of the material. If $q_e$ is the charge on the electron and $V_0$ is the stopping potential, then the work done by the retarding potential in stopping the electron is $q_e{V}_0$ , so we have $q_e{V}_0=K_{max}$. For hydrogen discharge tube: $hf=13.6eV$. In order to reduce the photoelectric current to zero, the voltage of anode relative to cathode must be $V=(3.8-13.6)eV=-9.8eV$ , where, the work function of the cathode surface is $3.8eV.$

So, the answer is option (B).