Question

If the frequency of the incident radiation is increased from $4 \times {10^{15}}Hz$ to $8 \times {10^{15}}Hz$ , by how much will the stopping potential for a given photosensitive surface go up?

Answer 1

Hint: In this question, we need to evaluate the amount of the stopping potential required for the given photosensitive surface such that the frequency of the incident radiation increases from $4 \times {10^{15}}Hz$ to $8 \times {10^{15}}Hz$. For this, we will use Einstein's photoelectric effect equation for this question by substituting the values from the two equations obtained accordingly.

Complete step by step answer:
Given that the frequency of the incident radiation is increased from $4 \times {10^{15}}Hz$ to $8 \times {10^{15}}Hz$
When light is incident on a metal surface, above a minimum energy barrier, the electrons are ejected with zero or some kinetic energy. This effect is known as the photoelectric effect. The potential to be applied to stop these kinetically active electrons is called stopping potential.
Einstein’s photoelectric effect can be, mathematically expressed as
Where $h$ is the planck's constant, $\nu $ is the frequency of incident radiation, $\phi $ is the work function, ${V_o}$ is the stopping potential and $e$ be the charge of the electron
Let ${\lambda _1} = 4 \times {10^{15}}Hz$ and ${\lambda _1} = 8 \times {10^{15}}Hz$
Then the corresponding stopping potential will be ${V_o}^1$ and ${V_o}^2$
Now we write the equations of photoelectric effect for both the cases.
$h{\nu _1} = \phi + e{V_o}^1$
And
$h{\nu _2} = \phi + e{V_o}^2$
Subtracting both we get
$h\left( {{\nu _1} - {\nu _2}} \right) = e\left( {{V_o}^1 - {V_o}^2} \right)$
Now we put the values, given in the question, and the constants and finally calculate the answer.
$
  V = \dfrac{{6.626 \times {{10}^{ - 34}} \times 4 \times {{10}^{15}}}}{{1.6 \times {{10}^{ - 19}}}} \\
\Rightarrow V = \dfrac{{6.626 \times 4}}{{1.6}} \\
\therefore V = 16.565V \\
 $
Thus the change in stopping potential is $16.565V$

Note:
Einstein’s photoelectric effect equation is a very important equation in physics. Approximation techniques used in big calculations should be known to the candidates as calculations are raised to certain raised powers. Candidates should also think of what factors do the stopping potential depend on and then write the mathematical expression accordingly.