Question

An electromagnetic radiation of frequency $3 \times {10^{15}}$ Hz falls on a photo surface whose work function is 4 eV. Then the maximum velocity of the photoelectrons emitted from the surface is
(given $h = 6.6 \times {10^{ - 34}}Js,m = 9 \times {10^{ - 31}}Kg$ )
A. $1.7 \times {10^6}m{s^{ - 1}}$
B. $3.4 \times {10^6}m{s^{ - 1}}$
C. $2.5 \times {10^6}m{s^{ - 1}}$
D. $2.0 \times {10^6}m{s^{ - 1}}$

Answer 1

Hint: When light (electromagnetic radiation) falls on a surface, some energy is required to release photons from the surface and the rest is used to increase the K.E. of the ejected photons. According to photoelectric effect, $K.E = E + {E_o} = h\upsilon - h{\upsilon _o}$ , where $h{\upsilon _o}$ is the energy of the incident photons (Work function), $h\upsilon $ is the energy of ejected photons. Substitute the values and find the value of K.E. Kinetic energy is also given as $\dfrac{1}{2}m{v^2}$ , where mass is given. Substitute the values of mass and obtained K.E. and find the value of the velocity v, where v is the velocity of the emitted photons.

Complete step by step answer:
We are given an electromagnetic radiation of frequency $3 \times {10^{15}}$ Hz falls on a photo surface whose work function is 4 eV.
And we have to calculate the maximum velocity of the photoelectrons emitted from the surface.
$K.E = E + {E_o} = h\upsilon - h{\upsilon _o}$, where h is the Planck’s constant, $E$ is the energy of the ejected photons from the photo surface and ${E_o}$ is the energy of the incident photons to the photo surface which can also be called as Wok function.
$
  h{\upsilon _o} = W \\
  K.E = h\upsilon - W \\
  W = 4eV,h = 6.6 \times {10^{ - 34}}Js,\upsilon = 3 \times {10^{15}}Hz \\
  K.E = \left( {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^{15}}} \right) - 4eV \\
  1eV = 1.6 \times {10^{ - 19}}J \\
\implies K.E = \left( {6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^{15}}} \right) - \left( {4 \times 1.6 \times {{10}^{ - 19}}} \right) \\
\implies K.E = \left( {19.8 \times {{10}^{ - 19}}} \right) - \left( {6.4 \times {{10}^{ - 19}}} \right) \\
\implies K.E = 13.4 \times {10^{ - 19}} = 1.34 \times {10^{ - 18}}J \\
 $
Kinetic energy is also equal to $\dfrac{1}{2}m{v^2}$
$
  K.E = \dfrac{1}{2}m{v^2} \\
  v = \sqrt {\dfrac{{2K.E}}{m}} \\
  m = 9 \times {10^{ - 31}}Kg,K.E = 1.34 \times {10^{ - 18}}J \\
  v = \sqrt {\dfrac{{2 \times 1.34 \times {{10}^{ - 18}}}}{{9 \times {{10}^{ - 31}}}}} \\
\implies v = \sqrt {2.97 \times {{10}^{12}}} \\
\therefore v = 1.7 \times {10^6}m/s \\
 $
The maximum velocity of the photoelectrons emitted from the surface is $1.7 \times {10^6}m{s^{ - 1}}$

So, the correct answer is “Option A”.

Note:
The above given question describes a phenomenon called photoelectric effect, when light hits a surface, it ejects electrons from them and the photon delivers its total energy to a single electron; whereas Compton Effect is another phenomenon, the photon transfers part of its energy to a single electron. The definitions of these two phenomena are almost the same but they both are completely different. So be careful while defining photoelectric effects.