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 photocurrent to zero the voltage of the anode relative to the cathode must be made

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 photocurrent 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

Tardigrade Admin · Accepted Answer

E=W0+eV0 For hydrogen atom, E=+13.6 eV ∴ +13.6=4.2+eV0 ⇒ V0=((. 13 . 6 - 4 . 2 .) eV/e)=9.4 V Potential at anode =-9.4 V

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

$- 4.2 V$

$- 9.4 V$

$- 17.8 V$

$+ 9.4 V$

$E = W_{0} + e V_{0}$
For hydrogen atom, $E = + 13.6 e V$
$∴ + 13.6 = 4.2 + e V_{0}$
$\Rightarrow V_{0} = \frac{( 13.6 - 4.2 ) e V}{e} = 9.4 V$
Potential at anode $= - 9.4 V$

Q. 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 reduce the photocurrent to zero the voltage of the anode relative to the cathode must be made

−4.2V

−9.4V

−17.8V

+9.4V