Question

The threshold frequency for a certain metal is $v_0$. When light of frequency $2v_0$ is incident on it, the maximum velocity of photoelectrons is $4\times 10^6 ms^{-1}$. If the frequency of incident radiation is increased to $5v_0$, then the maximum velocity of photoelectrons will be:
A. $4/5 \times 10^6\; ms^{-1}$
B. $2\times 10^6\; ms^{-1}$
C. $8\times 10^6\; ms^{-1}$
D. $2\times 10^7\; ms^{-1}$

Answer 1

Hint: Two cases have been given in this question with two different frequency values of incident radiation. We will use Einstein's photoelectric equation which gives us a relation between the kinetic energy of a photoelectron when a radiation having energy more than its threshold energy falls on it. With this, we will find the value of unknowns in the first case and then will utilize it to get the velocity in the second case.

Formula used: Einstein’s photoelectric equation, $E_k = hv- \phi$, where $E_k$ is the energy of emitted electron, h is the Planck’s constant, $v$ is the frequency of radiation and $\phi$ is the threshold energy of the electron.

Complete step by step answer:
We have been given that the frequency of incident light initially is $2v_0$.
The threshold frequency of the metal is $v_0$, and the maximum velocity of the electrons, $v_1 = 4\times 10^6\; ms^{-1}$,
Since, the energy from the incident ray will be utilized to remove the electrons from the surface and the remaining energy will get converted to kinetic energy of the electron.
So, by using Einstein’s photoelectric equation, $E_k = hv- \phi$ ………. (i)
where $E_k$ is the energy of an emitted electron, h is the Planck’s constant, $v$ is the frequency of radiation and $\phi$ is the threshold energy of the electron.

We can write that
$2hv_0 =hv_0 +\dfrac{1}{2}mv_{1}^{2}$
$\implies hv_0 =\dfrac{1}{2}m\times (4\times 10^6)^2 \implies \dfrac{hv_0}{m} =8\times 10^{12}$ ………. (ii)

So, now the frequency of incident radiation has been increased to a value of $5v_0$ and let the maximum velocity of photoelectrons now be $v_2$

Again, using the Einstein’s photoelectric equation from equation (i), we get

$5hv_0 = hv_0 + \dfrac{1}{2}m {v}_{2}^{2}$
$\implies \dfrac{8hv_0}{m}=v_2^{2}$
Substituting equation (ii), we get

$\implies v_2^{2} = 8 \times 8\times 10^{12} \implies v_2=8\times 10^6 ms^{-1}$
Hence, option C is the correct answer.

Note:
It should be noted that the complete energy of incident radiation will not get converted to the kinetic energy of the electron and require a threshold of energy to break its binding force from the atom or molecule. Sometimes, the work function is also known as the work function of the metal.