Question

The threshold frequency for certain metals is \[{{v}_{0}}\]. If the light of frequency \[2{{v}_{0}}\] is incident on it, electrons comes out with the maximum velocity of \[4\times {{10}^{6}}m/s\]. If the frequency of \[5{{v}_{0}}\]is incident on it, then the maximum velocity of photoelectrons will be:
A. \[8\times {{10}^{6}}m/s\]
B. \[16\times {{10}^{6}}m/s\]
C. \[2\times {{10}^{6}}m/s\]
D. \[12\times {{10}^{6}}m/s\]

Answer 1

Hint: In this question we have been asked to calculate the maximum velocity of photoelectrons if the frequency incident on it is increased to \[5{{v}_{0}}\]. We have been given the velocity of the photoelectrons. Therefore, to solve this question, we shall use Einstein’s equation of the photoelectric effect. This equation gives the relation between the energy of photoelectrons, the work is done at threshold frequency, and their kinetic energy.
Formula used:- \[K.E=E-W\]
Where,
K.E is the kinetic energy
E is the total energy
And W is the energy at the threshold frequency

Complete step-by-step solution
Einstein assumed that light is a particle and the energy carried by each particle is given by,
\[E=hv\] …………….. (1)
Where, \[v\] is the frequency of light and h is the Planck’s constant
The work function at threshold frequency is given by,
\[W=h{{v}_{0}}\] …………….. (2)
We know that,
\[K.E=\dfrac{1}{2}m{{v}^{2}}\] ……………….. (3)
Now, from Einstein’s equation of photoelectric effect
We know,
\[K.E=E-W\] ……………… (4)
Now, let us consider the first case where the frequency of incident light is given to be \[2{{v}_{0}}\]
Now, from (1), (2) and (3) the equation (4) can be written as,
\[\dfrac{1}{2}mv_{1}^{2}=h(v-{{v}_{0}})\]
After substituting the values
We get,
\[\dfrac{1}{2}mv_{1}^{2}=h(2{{v}_{0}}-{{v}_{0}})\]
Therefore,
\[\dfrac{1}{2}mv_{1}^{2}=h{{v}_{0}}\] …………………… (A)
Similarly,
For the second case, when the frequency of incident light is increased to \[5{{v}_{0}}\]
We can write,
\[\dfrac{1}{2}mv_{2}^{2}=h(5{{v}_{0}}-{{v}_{0}})\]
Therefore,
\[\dfrac{1}{2}mv_{2}^{2}=4h{{v}_{0}}\] ………… (B)
Now from (A) and (B)
We get,
\[\dfrac{1}{2}mv_{2}^{2}=4\dfrac{1}{2}mv_{1}^{2}\]
On solving,
We get,
\[{{v}_{2}}=2{{v}_{1}}\]
It is given that,
\[{{v}_{1}}=4\times {{10}^{6}}m/s\]
So, we get,
\[{{v}_{2}}=2\times 4\times {{10}^{6}}m/s\]
Therefore,
\[{{v}_{2}}=8\times {{10}^{6}}m/s\]
Therefore, the correct answer is option A.

Note: When electromagnetic radiation such as light hits the surface of the material, the electrons from the metal are emitted. This phenomenon is known as the photoelectric effect. The emitted electrons are known as photoelectrons. The photon beam has characteristic energy known as photon energy. When the electron in metal is hit by such a photon, it acquires the photon energy. If the photon energy acquired by the electron is greater than the binding energy, the electron is ejected.