Question

When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photocurrent is $\dfrac{{{V_0}}}{2}$. When the surface is illuminated by monochromatic light of frequency $\dfrac{v}{2}$, the stopping potential is ${V_0}$. The threshold frequency for photoelectric emission is:
A. $\dfrac{{3v}}{2}$
B. $2v$
C. $\dfrac{{4v}}{3}$
D. $\dfrac{{5v}}{3}$

Answer 1

Hint: In this question, we need to determine the threshold frequency for photoelectric emission such that monochromatic light of different frequencies falls on the photosensitive surface, releasing photoelectrons. For this we will use the relation of the photoelectric effect.

Complete step by step answer:
The energy of the photo electrons which are emitted from the photosensitive surface will always remain constant irrespective of the frequency of the monochromatic light rays falling on its surface.
Let W be the energy of the photoelectrons emitted from the photosensitive surface.
According to photo electric effect, the total energy of the emitted electrons will be the sum of the fixed potential energy and the variable kinetic energy. Mathematically, $E = W + eV$ where, E is the total energy, W is the constant threshold energy and ‘eV’ is the variable kinetic energy. This equation can be further written as $hv = W + eV$
According to the question, when a monochromatic light of frequency v, the stopping potential for the photocurrent is $\dfrac{{{V_0}}}{2}$. So, substituting the values in the equation $hv = W + eV$, we get
$
  hv = W + eV \\
   \Rightarrow hv = W + \dfrac{{e{V_0}}}{2} - - - - (i) \\
 $
Similarly, when a monochromatic light of frequency $\dfrac{v}{2}$, the stopping potential is ${V_0}$. So, substituting the values in the equation $hv = W + eV$, we get
$
  hv = W + eV \\
   \Rightarrow \dfrac{{hv}}{2} = W + e{V_0} - - - - (ii) \\
 $
Solving the equation (i) and (ii) to determine the value of the threshold energy.
From the equation (ii), we can write
$
  \dfrac{{hv}}{2} = W + e{V_0} \\
   \Rightarrow e{V_0} = \dfrac{{hv}}{2} - W - - - (iii) \\
 $
Substituting the value from the equation (iii) in the equation (i), we get
\[
  hv = W + \dfrac{{e{V_0}}}{2} \\
   \Rightarrow hv = W + \dfrac{{\left( {\dfrac{{hv}}{2} - W} \right)}}{2} \\
   \Rightarrow hv = W + \dfrac{{hv}}{4} - \dfrac{W}{2} \\
   \Rightarrow W - \dfrac{W}{2} = hv - \dfrac{{hv}}{4} \\
   \Rightarrow \dfrac{W}{2} = \dfrac{{3hv}}{4} \\
   \Rightarrow W = \dfrac{{3hv}}{2} \\
 \]
Hence, the threshold energy is \[\dfrac{{3hv}}{2}\]. Also, the threshold energy is the product of the Planck’s constant and the threshold frequency. So,
\[
  h{v_0} = \dfrac{{3hv}}{2} \\
   \therefore {v_0} = \dfrac{{3v}}{2} \\
 \]
Hence, the threshold frequency for photoelectric emission is $\dfrac{{3v}}{2}$

So, the correct answer is “Option A”.

Note:
According to photoelectric effect, whenever a light ray falls on the photo-sensitive surface then the photoelectrons get emitted through the surface. But the light ray should be such that it reaches the threshold energy of the surface.