Question

If the work function of a metal is 3.6eV where it is connected as an emitter in a photoelectric experiment and light with wavelength 732nm strikes metal. How to find if there is minimum potential difference that can be applied to ensure that there is no current in the circuit due to the movement of photoelectrons to the collector?

Answer 1

For no current in the circuit due to the movement of photoelectrons, the velocity of emitted photoelectrons should be equal to zero

$KE=\dfrac{hc}{\lambda}-W_o=0$

$\therefore\dfrac{hc}{\lambda}=W_o$

or

$\dfrac{hc}{\lambda}<W_o$

In our case,

$\lambda=732\space nm$

$W_o=3.6 \space eV$

$\dfrac{hc}{\lambda}=[\dfrac{(6.63\times10^{-34})(3\times10^8)}{732\times10^{-9}}\times\dfrac{1}{1.6\times10^{-19}}]\space eV$

$\dfrac{hc}{\lambda}=1.69\space eV$

Since,

$\dfrac{hc}{\lambda}<W_o$

Therefore, no photon is emitted from the surface of the metal and no current flows in the circuit.

Minimum voltage required = 3.6 V