Question

Given that a photon of light of wavelength \[10,000\overset{\text{o}}{\mathop{\text{A}}}\,\]has an energy equal to \[1.23\text{ eV}\]. When light of wavelength \[5000\overset{\text{o}}{\mathop{\text{A}}}\,\]and intensity \[{{I}_{\text{o}}}\]falls on a photoelectric cell, the saturation current is \[0.40\times {{10}^{-6}}\text{ A}\] and the stopping potential is \[1.36\text{ V}\]; then the work function is:
(A) \[0.43\text{ eV}\]
(B) \[1.10\text{ eV}\]
(C) \[1.36\text{ eV}\]
(D) \[2.47\text{ eV}\]

Answer 1

Hint:
 The maximum kinetic energy is independent of intensity of light and also the saturation current. The work function depends on the wavelength incident light and the stopping potential of the photoelectric cell.
Formula used:
The work-function W of the metal in terms of maximum kinetic energy\[{{E}_{k}}\] is given by:
\[W=\dfrac{hc}{\lambda }-{{E}_{k}}\]
Where \[\lambda \] is the wavelength of incident radiation, c is the speed of light and h is the Planck’s constant.
Now, the stopping potential \[{{V}_{s}}\]is a measure of the maximum kinetic energy \[{{E}_{k}}\]of the electrons and is therefore given by:
\[{{E}_{k}}=e{{V}_{s}}\]
Where e is the charge on electrons.

Complete step by step answer:
Wavelength of incident light, \[\lambda =5000\overset{\text{o}}{\mathop{\text{A}}}\,=5000\times {{10}^{-10}}\text{ m}\]
The stopping potential, \[{{V}_{s}}=1.36\text{ V}\]
Speed of light in air, \[c=3\times {{10}^{8}}\text{ m/s}\]
Planck’s constant, \[h=6.6\times {{10}^{-34}}\text{ Js}\]
The charge on electrons, \[e=1.6\times {{10}^{-19}}\text{ C}\]
Substitute the value of e and \[{{V}_{s}}\], in the kinetic energy-formula to get maximum kinetic energy \[{{E}_{k}}\]:
\[\begin{align}
& {{E}_{k}}=(1.6\times {{10}^{-19}}\text{ C)(1}\text{.36 V)} \\
& {{E}_{k}}=2.176\times {{10}^{-19}}\text{ J} \\
\end{align}\]
Now, substitute the values of h, c, \[\lambda \] and \[{{E}_{k}}\] in the work function-formula to get W:
\[\begin{align}
& W=\dfrac{hc}{\lambda }-{{E}_{k}} \\
& W=\dfrac{(6.6\times {{10}^{-34}}\text{ Js})(3\times {{10}^{8}}\text{ m/s})}{5000\times {{10}^{-10}}\text{ m}}-(2.176\times {{10}^{-19}}\text{ J)} \\
& W=(3.96\times {{10}^{-19}}\text{ J)}-(2.176\times {{10}^{-19}}\text{ J)} \\
& W=1.78\times {{10}^{-19}}\text{ J} \\
& W=1.10\text{ eV }\!\![\!\!\text{ 1 eV}=1.6\times {{10}^{-19}}\text{ J }\!\!]\!\!\text{ } \\
\end{align}\]
Therefore the work function of the photoelectric cell is \[1.10\text{ eV}\].
Hence, option B is the correct answer.
Additional information:
The minimum energy required for the emission of photoelectrons from a metal is called the ‘work function’ of the metal.

Note: All the physical quantities should be in the same unit system, preferably the S.I. unit system. Convert the wavelength given in angstrom to meter before calculating work function in joules. Finally convert the work function obtained in joules to electron-volt, to get the desired answer; else the answer will not match with the options. The intensity of light, saturation current of the cell are extra information that are not utilised to solve the problem.