Question

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second light source of wavelength l/2 ejects photoelectrons with a maximum kinetic energy of 4 eV. What is the work function of the metal

Answer 1

Explanation:

From first source, kinetic energy ($K.E_{1}$) ejected is 1 eV and wavelength of light is $\lambda$.

From second source, kinetic energy ($K.E_{2}$) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$.

Relation between work function, wavelength, and kinetic energy is as follows.

                   K.E = $\frac{hc}{\lambda} - \phi$

where,        h = Plank's constant = $6.63 \times 10^{-34}$ J.s

                   c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

• From first source,

                      $K.E_{1}$ = $\frac{hc}{\lambda} - \phi$  

            1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$    

    $1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$        ........... (1)

• From second source,

                  $K.E_{2}$ = $\frac{hc}{\lambda} - \phi$  

          $4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$        

                 $6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$        ........... (2)        

Now, divide equation (2) by 2. Therefore, it will become

       ${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

                $3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$   ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

                 $1.6 \times 10^{-19} = \frac{\phi}{2}$

                    $\phi$ = $3.2 \times 10^{-19}$

                          = 2 eV

Thus, we can conclude that work function of the metal is 2 eV.