Light from a hydrogen 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

Options

A. `+9.4` V
B. `-17.8` V
C. `-4.2` V
D. `-9.4` V

Answer

D

Explanation

For hydrogen atom, the energy of the light `E=13.6` eV
`:'E=W_(0)+K_("max")=W_(0)+eV_(0)`
`:.13.6=4.2+eV_(0)`
`:.V_(0)=((13.6-4.2)eV)/(e)=9.4 V`
Stopping potential `=9.4 V`
`:.` Potential at the anode `=-9.4V`