Question

Electromagnetic radiation having $\lambda = 310{\text{ }}{{\text{A}}^ \circ }$ is subjected to a metal sheet having work function $= 12.8{\text{ eV}}$. What will be the velocity of photo electrons having maximum kinetic energy?
A. ${\text{1}}{\text{.23}} \times {10^6}{\text{ m/s}}$
B. $2.28 \times {10^6}{\text{ m/s}}$
C. $3.1 \times {10^6}{\text{ m/s}}$
D. None of these

Answer 1

In this question, as the electromagnetic radiation hits the metal sheet photoelectrons will be ejected giving rise to the photoelectric effect. We can find the energy of the incident photons using the formula $E = hv$. For the photoelectric effect to take place energy of incident photons must be greater than the work function of the metal surface and the maximum kinetic energy of the photoelectrons is given by $K.E = E - W$. Using this equation, we can find out the maximum velocity.

Formula used:
Energy of a photon,
$E = hv = \dfrac{{hc}}{\lambda }$
Where,$h$= Planck’s constant = $6.63 \times 10^{-34}\,Js$ and $\lambda$ is the wavelength of the incident photon.
$K.{E_{\max }} = \dfrac{1}{2}m{{\text{v}}^2} = hv - W$
Where $hv$ is the energy of the incident photon and $W$ is the work function of the metal.

Complete step by step answer:
The wavelength of the incident photon is $\lambda = 310{\text{ }}{A^ \circ }$.
The work function of the metal surface is $W = 12.8{\text{ eV}}$.
Therefore, we can find energy of the incident photon using the formula,
$E = hv = \dfrac{{hc}}{\lambda }$
Substituting the values we get,
${\text{E = }}\dfrac{{\left( {6.62 \times {{10}^{ - 34}}} \right)\left( {3 \times {{10}^8}} \right)}}{{\left( {310 \times {{10}^{ - 10}}} \right)}}J$
$\Rightarrow E = \dfrac{{19.86 \times {{10}^{ - 17}}}}{{31}}$

On solving this we get,
The Energy of incident photon is $E = 6.406 \times {10^{ - 18}}J$
The Work function of the metal surface is,
$W = 12.8{\text{ eV = 12}}{\text{.8}} \times {\text{1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 19}}J$
$\Rightarrow W = 2.016 \times {10^{ - 18}}J$
The maximum kinetic energy of the photoelectrons ejected from the metal surface can be calculated using $K.{E_{\max }} = hv - W$
To find out the maximum velocity substitute,
$K.{E_{\max }} = \dfrac{1}{2}m{{\text{v}}^2}$

Thus, $\dfrac{1}{2}m{{\text{v}}^2} = hv - W$
Substituting, $W = 2.016 \times {10^{ - 18}}J$ , $E = 6.406 \times {10^{ - 18}}J$ and $m = 9.1 \times {10^{ - 31}}{\text{ kg}}$ in the equation we get,
$\dfrac{1}{2} \times \left( {9.1 \times {{10}^{ - 31}}{\text{ }}} \right){{\text{v}}^2} = \left( {6.406 \times {{10}^{ - 18}}} \right) - \left( {2.016 \times {{10}^{ - 18}}} \right)$
$\Rightarrow \dfrac{1}{2} \times \left( {9.1 \times {{10}^{ - 31}}{\text{ }}} \right){{\text{v}}^2} = 4.39 \times {10^{ - 18}}$
$\Rightarrow {{\text{v}}^2} = \dfrac{{4.39 \times {{10}^{ - 18}} \times 2}}{{9.1 \times {{10}^{ - 31}}}}$

On solving we get,
$\Rightarrow {{\text{v}}^2} = 9.648 \times {10^{12}}$
Taking root on both the sides
$\Rightarrow {\text{v}} = \sqrt {\left( {9.648 \times {{10}^{12}}} \right)}$
$\therefore {\text{v}} = 3.106 \times {10^6}{\text{ m/s}}$
Therefore, velocity of the photoelectrons having maximum kinetic energy is ${\text{v}} = 3.106 \times {10^6}{\text{ m/s}}$.

Therefore, option C is the correct answer.

Note: According to wave theory, the photoelectric effect should occur for any frequency of the light, provided that the light is intense enough. However, the equation$K.{E_{\max }} = hv - h{v_0}$ suggests that photoemission is possible only when the frequency of incident light is either greater than or equal to the threshold frequency ${v_0}$.