The work function for sodium and copper are 2eV and 4eV. Which of them is suitable for a photocell with \[{\rm{4000}}\mathop {\rm{A}}\limits^{\rm{0}} \] light?
A. Copper
B. Sodium
C. Both
D. Neither of them

Hint:Work function is the minimum energy required to remove one electron from a metal surface. Work function can also be said as the characteristics of the surface and not the complete metal. As we know that the energy of the incident light or photons should be more than the work function of the metal for the photoemission to take place.

Formula Used:
The formula to find the threshold wavelength is,
\[\lambda = \dfrac{{hc}}{W}\]
Where, h is Planck’s constant, c is the speed of light and W is a work function.

Complete step by step solution:
Here the work function for sodium and copper are given as 2eV and 4eV. Now, we need to find which of them is suitable for a photocell with \[{\rm{4000}}\mathop {\rm{A}}\limits^{\rm{0}} \] light. In order to find this, we have the formula for the threshold wavelength for sodium which is given as,
\[{\lambda _{Na}} = \dfrac{{hc}}{W}\]………….. (1)
Here, \[h = 6.626 \times {10^{ - 34}}Js\], \[c = 3 \times {10^8}m{s^{ - 1}}\]and \[W = 2eV\]

Since the work function is in electron volt. We need to convert Planck’s constant also from joules to eV. Then,
\[h = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{1.6 \times {{10}^{ - 19}}}}eV\]
\[\Rightarrow h = 4.14125 \times {10^{ - 15}}eV\]
Substitute the value of h, c and W in equation (1) we get,
\[{\lambda _{Na}} = \dfrac{{4.14125 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{2}\]
\[\Rightarrow {\lambda _{Na}} = 6.211 \times {10^{ - 7}}m\]
\[\Rightarrow {\lambda _{Na}} = 6211\mathop {\rm{A}}\limits^{\rm{0}} \]

Also, for copper we have,
\[{\lambda _{Cu}} = \dfrac{{hc}}{W}\]…….. (2)
Here, \[h = 6.626 \times {10^{ - 34}}Js\]
\[ \Rightarrow h = 4.14125 \times {10^{ - 15}}eV\],
\[\Rightarrow c = 3 \times {10^8}m{s^{ - 1}}\] and \[W = 4eV\]
Substitute the value of h, c and W in equation (2) we get,
\[{\lambda _{Cu}} = \dfrac{{4.14125 \times {{10}^{ - 15}} \times 3 \times {{10}^8}}}{4}\]
\[\Rightarrow {\lambda _{Cu}} = 3.105 \times {10^{ - 7}}m\]
\[\therefore {\lambda _{Cu}} = 3105\mathop {\rm{A}}\limits^{\rm{0}} \]
Therefore, \[{\lambda _{Na}} > 4000\mathop {\rm{A}}\limits^{\rm{0}} \], sodium is suitable for a photocell.

Hence, option B is the correct answer.

Note: Remember that, Since the work function is in electron volts and if the energy is given in joules, then we need to convert the given value of energy from joules to eV by dividing it by the charge of an electron while solving this problem.