Question

Direction: The question has Statement 1 and Statement 2. Of the four choices given after the statements, choose one that describes the two statements.
Statement 1: A metallic surface is irradiated by monochromatic light of frequency $v > {v_o}$ (the threshold frequency). The maximum kinetic energy and stopping potential are ${K_{\max }}$ and ${V_o}$ respectively. If the frequency incident on the surface is doubled, both the ${K_{\max }}$ and ${V_o}$ are also doubled
Statement 2: The maximum kinetic energy and stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
(A) Statement 1 is True, Statement 2 is False.
(B) Statement 1 is True, Statement 2 is True; Statement 2 is the correct explanation for Statement 1.
(C) Statement 1 is True, Statement 2 is True; Statement 2 is not the correct explanation for Statement 1.
(D) Statement 1 is False, Statement 2 is True.

Answer 1

From the photoelectric equation we can write the kinetic energy of the photoelectrons is the difference of the energy of the photons and the work function. The kinetic energy is equal to the product of the e and the stopping potential. From there we can check if the frequency is doubled, whether the kinetic energy also gets doubled.
In this solution we will be using the following formula,
$\Rightarrow hv = \phi + K.E.$
where $hv$ is the energy of the photons,
$\phi$ is the work function of the metal
and $K.E.$ is the kinetic energy of the photo electrons.

Complete step by step solution:
It is given in the question that a light is incident on the surface of the metal which has a frequency $v > {v_o}$. Therefore, from Einstein’s photoelectric equation we can write,
$\Rightarrow hv = \phi + K.E.$
We can write this in the terms of kinetic energy as,
$\Rightarrow K.E. = hv - \phi$
Now when the frequency of the light is doubled, we have the kinetic energy of the photoelectrons as,
$\Rightarrow K.E.' = 2hv - \phi$
The work function doesn’t change because it is the property of the metal. So if we take the ratio of the two kinetic energies, we get
$\Rightarrow \dfrac{{K.E.'}}{{K.E.}} = \dfrac{{2hv - \phi }}{{hv - \phi }}$
We can see that the value of this ratio is clearly not equal to 2.
Again, the kinetic energy can be written in the terms of stopping potential as,
$\Rightarrow K.E. = e{V_o}$
So we can also write, $e{V_o} = hv - \phi$
Again on doubling the frequency of the incident light we have,
$\Rightarrow e{V_o}^\prime = 2hv - \phi$
Taking the ratio of the two cases we get,
$\Rightarrow \dfrac{{e{V_o}^\prime }}{{e{V_o}}} = \dfrac{{2hv - \phi }}{{hv - \phi }}$
Here we can cancel the e from the LHS and get,
$\Rightarrow \dfrac{{{V_o}^\prime }}{{{V_o}}} = \dfrac{{2hv - \phi }}{{hv - \phi }}$
Here also we can see that when the frequency gets doubled, the stopping potential doesn’t become two times of its original value. Hence the statement 1 is false.
Now from the equations $K.E. = hv - \phi$ and $e{V_o} = hv - \phi$, since the work function is constant in both the equations, the kinetic energy and the stopping potential are both directly proportional to the frequency of the light $v$.
So we can write$K.E. \propto v$ and ${V_o} \propto v$. Hence the kinetic energy and stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.

Therefore, the statement 1 is false and statement 2 is correct. So the correct answer will be option (D).

Note:
The photoelectric effect is the phenomenon where light is emitted from the surface of a metal, when the light of a certain frequency strikes on it. The electrons emitted in this process are called the photo electrons.