# The ejection of the photoelectron from the silver metal in the photoelectric

Question:

The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Solution:

From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e

$E=W_{0}+K \cdot E$

$\Rightarrow W_{0}=E-K \cdot E$

Energy of incident photon $(E)=\frac{h c}{\lambda}$

Where,

$c=$ velocity of radiation

h = Planck's constant

$\lambda=$ wavelength of radiation

Substituting the values in the given expression of $E$ :

$E=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{~ms}^{-1}\right)}{256.7 \times 10^{-9} \mathrm{~m}}$

$=7.744 \times 10^{-19} \mathrm{~J}$

$=\frac{7.744 \times 10^{-19}}{1.602 \times 10^{-19}} \mathrm{eV}$

$E=4.83 \mathrm{eV}$

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,

$K \cdot E=0.35 \mathrm{~V}$

$\mathrm{K} \mathrm{E}=0.35 \mathrm{eV}$

$\therefore$ Work function, $W_{0}=E-K \cdot E$

$=4.83 \mathrm{eV}-0.35 \mathrm{eV}$

$=4.48 \mathrm{eV}$