Question

Two photons of energy twice and thrice the work function of the metal is incident on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively is:
A.$\sqrt 2 :1$
B.$\sqrt 3 :1$
C.$\sqrt 3 :\sqrt 2$
D.$1:\sqrt 2$

Answer 1

When a metal surface is exposed to light rays of a sufficient amount of shorter wavelength, the incident radiation will be absorbed by the metal surface and this, in turn, emits electrons. This process is called the photoelectric effect and the emitted electrons are called photoelectrons. The emission of the electron varies directly as the intensity of the light falling on the surface. The work function of the material is defined as the minimum kinetic energy required to pull out the electrons from the surface.

Complete answer:
The work function can be represented as $\phi$. The value of the work function depends upon the metal.
It is said that the incident energy of the photon must be equal to the sum of the photoelectron’s kinetic energy and the work function of the metal.
${E_{photon}} = K{E_{electron}} + \phi$
This value of the kinetic energy is the maximum possible kinetic energy of the photoelectron. Einstein determined an equation, which says (maximum kinetic energy of an electron) = (energy of the incident light energy packet) minus (the work function)
Mathematically,
$K{E_{\max }} = {E_{photon}} - \phi$
Given in the question that two incident energy of the photon is twice and thrice of the work function of the metal.
So for the first photon, the incident energy is twice the work function. Therefore we can write it as,
$\dfrac{1}{2}m{v_1}^2 = 2\phi - \phi$
$\Rightarrow \dfrac{1}{2}m{v_1}^2 = \phi$
$\Rightarrow {v_1}^2 = \dfrac{{2\phi }}{m}$
$\Rightarrow {v_1} = \sqrt {\dfrac{{2\phi }}{m}}$ ……… (1)
For the second photon, the incident energy is twice the work function. Therefore we can write it as,
$\dfrac{1}{2}m{v_2}^2 = 3\phi - \phi$
$\Rightarrow \dfrac{1}{2}m{v_2}^2 = 2\phi$
$\Rightarrow {v_2}^2 = \dfrac{{4\phi }}{m}$
$\Rightarrow {v_2} = \sqrt {\dfrac{{4\phi }}{m}}$ …….. (2)
The ratio between the two maximum velocities can be written as,
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{\sqrt {\dfrac{{2\phi }}{m}} }}{{\sqrt {\dfrac{{4\phi }}{m}} }} = \dfrac{{1}}{{\sqrt 2 }}$

Therefore the correct option is D.

Note:
The maximum kinetic energy is independent of the intensity of the light falling on the metal surface. This intensity is the energy of the light falling per unit area, per unit time. Also, this maximum kinetic energy increases linearly with the frequency of the light and below a certain frequency, there will be no emission of photoelectrons.