Light of wavelength \[1824\mathop A\limits^ \circ \], incident on the surface of a metal produces photoelectrons with maximum energy 5.3 eV. Now, when the light of wavelength \[1216\mathop A\limits^ \circ \] is used, the maximum energy of photoelectrons is 8.7 eV. The work function of the metallic surface is:
A. 6.8 eV
B. 1.5 eV
C. 13.6 eV
D. 3.5 eV

Hint: The maximum kinetic energy of the photoelectron is calculated by finding the difference of the total energy of the photon and the work function of the metal. When the energy of the photon and maximum kinetic energy of the photoelectron is given then we calculate the work function from this and use it to find other quantities.

Formula used:
\[E = \dfrac{{hc}}{\lambda } - {\phi _0}\]
here E is the kinetic energy of the emitted electron, h is the planck's constant, c is the speed of light, \[\lambda \] is the wavelength of the photon and \[{\phi _0}\] is the work function of the metal.

Complete step by step solution:
When light of wavelength \[1824\mathop A\limits^ \circ \] is incident on the surface of a metal then it produces photoelectrons with maximum energy 5.3 eV.
\[5.3eV = \dfrac{{eV}}{{1.6 \times {{10}^{ - 34}}}}\left( {\dfrac{{hc}}{{1.824 \times {{10}^{ - 7}}}}} \right) - {\phi _0} \\ \]
\[\Rightarrow 5.3eV = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}\left( {\dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{1.824 \times {{10}^{ - 7}}}}} \right)eV - {\phi _0} \\ \]
\[\Rightarrow 5.3eV = 6.811eV - {\phi _0} \\ \]
\[\Rightarrow {\phi _0} = \left( {6.811 - 5.3} \right)eV \\ \]
\[\Rightarrow {\phi _0} = 1.51eV \ldots \left( i \right)\]

Now, when the light of wavelength \[1216\mathop A\limits^ \circ \]is used, the maximum energy of photoelectrons is 8.7 eV.
\[8.7eV = \dfrac{{eV}}{{1.6 \times {{10}^{ - 34}}}}\left( {\dfrac{{hc}}{{1.216 \times {{10}^{ - 7}}}}} \right) - {\phi _0} \\ \]
\[\Rightarrow 8.7eV = \dfrac{1}{{1.6 \times {{10}^{ - 19}}}}\left( {\dfrac{{\left( {6.626 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{1.216 \times {{10}^{ - 7}}}}} \right)eV - {\phi _0} \\ \]
\[\Rightarrow 8.7eV = 10.22eV - {\phi _0} \\ \]
\[\Rightarrow {\phi _0} = \left( {10.11 - 8.7} \right)eV \\ \]
\[\therefore {\phi _0} = 1.52eV \ldots \left( {ii} \right)\]
From both the data it can be said that the work function of the given metal is approximately 1.5 eV.

Therefore,the correct option is B.

Note: To check the final answer we can verify by using in the first part of the question where the energy of the photon is given as well as the maximum energy. If the work function comes out the same in both parts then the final answer is correct.