Question

A monochromatic light of intensity $5mW$ emits $8\times {{10}^{15}}$ photons per second. The light ejects photoelectrons from a metal surface. The stopping potential is 2eV. The work function of that metal is:
$\begin{aligned} & (A)1.9eV \\\ & (B)2.9eV \\\ & (C)3.2eV \\\ & (D)1.2eV \\\ \end{aligned}$

Answer 1

Since, the power intensity of light and number of photons emitted at this intensity are provided. On dividing the power by the number of photons, we will get the energy per photon. We will then convert it into electron-volt. Then, finally using the formula for the work function of a metal, that is total energy minus the stopping potential, we can calculate the work function of the given metal.

Complete step-by-step answer:
Let the intensity of light be denoted by $P$. Then, its value has been given to us as:
$\Rightarrow P=5mW$
$\Rightarrow P=5\times {{10}^{-3}}W$ [Let this expression be equation number (1)]
Now, let the number of photons be denoted by $n$. Then, $n$is equal to:
$\Rightarrow n=8\times {{10}^{15}}$ [Let this expression be equation number (2)]
Thus, the total energy of one photon (say E) can be calculated as:
$\Rightarrow E=\dfrac{P}{n}$
That is, the division of equation number (1) by (2).
Putting the values of these terms in the above equation, we get:
$\begin{aligned} & \Rightarrow E=\dfrac{5\times {{10}^{-3}}}{8\times {{10}^{15}}}J \\\ & \Rightarrow E=6.25\times {{10}^{-19}}J \\\ \end{aligned}$
Converting it into electron-volt, we get:
$\Rightarrow E=\dfrac{6.25\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV$
$\Rightarrow E=3.9eV$
Now, the work function of a metal is given by the following formula:
$\Rightarrow W=E-{{V}_{S}}$
Where,
E is the total energy of a photon which is calculated as 3.9eV
${{V}_{S}}$ is the stopping potential of the metal which is given as 2eV
Putting the values of these terms in the above equation, we get the work function as:
$\begin{aligned} & \Rightarrow W=\left( 3.9-2 \right)eV \\\ & \therefore W=1.9eV \\\ \end{aligned}$
Hence, the work function of the metal comes out to be 1.9eV .

So, the correct answer is “Option A”.

Note: In case of photoelectric phenomenon, if the total energy of incident light wave is less than the stopping potential of the metal, then no photoelectrons emit from the metal. Also, the most widely used element in photoelectric effect is Cesium because of its very low ionization energy.