Question

A light source of wavelength $ \lambda $ illuminates a metal and ejects photo electron with $ {\left( {K.E} \right)_{\max .}} = {\text{ 1 }}e.v $ . Another light source of wavelength $ \dfrac{\lambda }{3} $ ,ejects photo-electrons from same metal with $ {\left( {K.E} \right)_{\max .}} = {\text{ 4 }}e.v $ . Find the value of the work function of metal.

Answer 1

Hint: When the light source incident on a certain metal then the energy carried by the photons of light is used for the kinetic energy of electrons of the metal and the work function of the metal. Thus the total energy is equal to the sum of maximum kinetic energy and the work function. We will use this relation to find the work function of metal in the above question.
 $ E = \dfrac{{hc}}{\lambda } $
 E is the energy, $ h $ is Planck’s constant , $ c $ is the speed of light and $ \lambda $ is the wavelength of photons.

Complete answer:
When a light of source of energy, $ \left( E \right) $ is made to fall on certain metal then this energy is equal to sum of kinetic energy $ \left( {K.E} \right) $ of electrons of metals and work function $ \left( W \right) $ of that metal. It can be represented as:
 $ E = {\text{ }}K.E{\text{ + }}W $
Since we know that energy E can be written as:
 $ E = \dfrac{{hc}}{\lambda } $
Thus the above later equation can be reduced as:
 $ \dfrac{{hc}}{\lambda } = {\text{ }}K.E{\text{ + }}W $
Now according to question when wavelength of light source is $ \lambda $ then $ {\left( {K.E} \right)_{\max .}} = {\text{ 1 }}e.v $ which can be represented as:
 $ \dfrac{{hc}}{\lambda } = {\text{ 1 + }}W $ _________ $ (i) $
Also when wavelength of wave is $ \dfrac{\lambda }{3} $ then $ {\left( {K.E} \right)_{\max .}} = {\text{ 4 }}e.v $ which can be represented as:
 $ \dfrac{{hc}}{{\dfrac{\lambda }{3}}} = {\text{ 4 + }}W $
 $ \dfrac{{3hc}}{\lambda } = {\text{ 4 + }}W $
From equation $ (i) $ the above can be reduced by using $ \dfrac{{hc}}{\lambda } = {\text{ 1 + }}W $ :
 $ \Rightarrow 3\left( {1 + W} \right) = {\text{ 4 + }}W $
 $ \Rightarrow 3 + 3W = {\text{ 4 + }}W $
The equation can be rearranged and we get work function as:
 $ \Rightarrow W = {\text{ }}\dfrac{1}{2}{\text{ }}e.v $
Therefore the work function of the metal is $ \dfrac{1}{2}{\text{ }}e.v $ or $ 0.5{\text{ }}e.v $ .

Note:
There is no need to use the value of Planck’s constant and speed of light as we are going to substitute the value of $ \dfrac{{hc}}{\lambda } = {\text{ 1 + }}W $ . The energy which a photon carries first is used to overcome the work function of the metal and then rest of the energy is used for kinetic energy of the electrons of the metal.