Question

Photoelectric effect is observed from a metal surface when light of frequencies $3 \times {10^{14}}\;Hz$ and $2 \times {10^{14}}\;Hz$ are incident. If the maximum kinetic energies of the photoelectrons are in the ratio $2:1$ then for which of the following frequencies photoelectrons are emitted from the same metal plate?
A. $\dfrac{1}{2} \times {10^{14}}\;Hz$
B. $\dfrac{3}{2} \times {10^{14}}\;Hz$
C. $\dfrac{4}{3} \times {10^{14}}\;Hz$
D. $\dfrac{4}{5} \times {10^{14}}\;Hz$

Answer 1

: The condition for photoelectric effect to occur is that the frequency of light incident on it should be more than the threshold frequency of the metal or in other words the energy of incident light must have an energy more than that of the work function of the metal. The maximum kinetic energy of the emitted photons is determined by the difference between the energy of the ejected photon and the work function of the metal.

Complete step by step answer:
The phenomenon of emission of electrons from a metal surface when light is incident on it is called the photoelectric effect and this effect depends on the frequency of the incident light.We are required to find out which of the frequencies out of the given set of frequencies initiate the emission of photons.

This basically means that the value of the frequencies must be more than that of the threshold frequency so that photoelectric effect takes place. For this we first need to determine the value for threshold frequency. Threshold frequency is the frequency below which no photoelectrons are emitted, that is, when photoelectric effect does not take place.The work function of a metal is the minimum amount of energy required by an electron to just escape from its surface.

The first step in the process of calculating the threshold frequency is to obtain an equation relating the maximum kinetic energy of the photoelectrons and the work function of the metal in-order to construct equations using the data given in the problem. According to Einstein’s theory of photoelectric effect and by the principle of conservation of energy the equation is as follows:
$E = {K_{\max }} + \phi$
By rearranging the terms we get the maximum kinetic energy of the ejected photons to be:
${K_{\max }} = E - \phi$ ---($1$)
The work function of the metal is given by the formula:
$\phi = h{v_0}$ ---($2$)
Where, $\phi$ denotes the work function of a metal, $h$ denotes the Plank’s constant and ${v_0}$ denotes the threshold frequency.
Since $h$ is a constant they have values that have already been calculated. These value of Plank’s constant is:
$h = 6.626 \times {10^{ - 34}}\;Js$
Each photon emitted has a definite energy which depends on the frequency of the radiation.

The energy of the photons is given by the equation:
$E = hv$ ----($3$)
Where, $v$ denotes the frequency.
By substituting the equations ($3$) and ($2$) in ($1$) we get:
${K_{\max }} = hv - h{v_0}$
By keeping the above formula as a base let us construct an equation using the data given. The two frequencies of light are given. Let the two frequencies be ${v_1}$ and ${v_2}$. Also, let the maximum kinetic energies be ${K_1}$ and ${K_2}$. We get:
${K_{\max }} = h{v_1} - h{v_0}$
$\Rightarrow {K_{\max }} = h{v_2} - h{v_0}$
Given, ${v_1} = 3 \times {10^{14}}\;Hz$ and ${v_2} = 2 \times {10^{14}}\;Hz$.
It is given that we need to calculate for the same metal which means that the work function remains the same. The two equations after substituting the values will be as follows:
${K_1} = (h \times 3 \times {10^{14}}) - h{v_0}$ -----($4$)
$\Rightarrow{K_2} = (h \times 2 \times {10^{14}}) - h{v_0}$ -----($5$)

It is given in the question that the ratio between the two kinetic energies is $2:1$ and hence we can say:
$\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{2}{1}$
$\Rightarrow {K_1} = 2{K_2}$ -----($6$)
Dividing the equations ($4$) and ($5$) we get:
$\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{{(h \times 3 \times {{10}^{14}}) - h{v_0}}}{{(h \times 2 \times {{10}^{14}}) - h{v_0}}}$
Taking $h$ as common and substituting ($6$) in the above equation we get:
$\dfrac{{2{K_2}}}{{{K_2}}} = \dfrac{{h(3 \times {{10}^{14}} - {v_0})}}{{h(2 \times {{10}^{14}} - {v_0})}}$
Cancelling out the common terms we get:
$2 = \dfrac{{3 \times {{10}^{14}} - {v_0}}}{{2 \times {{10}^{14}} - {v_0}}}$
By cross multiplying we get:
$2(2 \times {10^{14}} - {v_0}) = 3 \times {10^{14}} - {v_0}$
By solving out the equation we get:
$4 \times {10^{14}} - 2{v_0} = 3 \times {10^{14}} - {v_0}$
$\Rightarrow 4 \times {10^{14}} - 3 \times {10^{14}} = 2{v_0} - {v_0}$
$\Rightarrow {v_0} = {10^{14}}(3 - 2)$
$\Rightarrow {v_0} = 1 \times {10^{14}}$

This is the threshold frequency which is the minimum frequency required for the effect to take place and hence we now compare each frequency given in the question with this threshold frequency. The frequency must be greater than the ${v_0}$ we have found out. From the give option:
Let, ${v_1} = \dfrac{1}{2} \times {10^{14}}$
$\Rightarrow {v_1} < {v_0}$
Let, ${v_2} = \dfrac{3}{2} \times {10^{14}}$
$\Rightarrow {v_2} \geqslant {v_0}$
Let, ${v_3} = \dfrac{4}{3} \times {10^{14}}$
$\Rightarrow {v_3} \geqslant {v_0}$
Let, ${v_4} = \dfrac{4}{5} \times {10^{14}}$
$\therefore {v_4} < {v_0}$
Hence we can see that the second and third frequencies are more than the threshold frequency which means that light of frequencies ${v_2}$ and ${v_3}$ initiate the photoelectric effect and the subsequent emission of electrons.

Therefore, option B and C are the correct answer.

Additional information: The principle for Einstein’s equation revolves around the concept of conservation of energy. A part of the energy of the photon is used in liberating an electron from the metal surface which is the work function of the metal and the remaining energy of the photon goes in imparting the kinetic energy to the ejected electron. The photoelectric effect is an instantaneous process and the time between the incidence of light and its emission is just ${10^{ - 9}}\;s$.

Note: The frequency of a photon does not change when it travels from one medium to another. Another alternative to find if the incident light allows photoelectric effect to take place is by calculating its energy and comparing with the work function of the metal and if it’s greater then also the condition satisfies.