Question

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda $, the stopping potential for photoelectric current is $3{{V}_{0}}$​ and when the same surface is illuminated with light of wavelength $2\lambda $, the stopping potential is ${{V}_{0}}$​. The threshold wavelength of this surface for photoelectric effect is
$\begin{align}
  & A)6\lambda \\
 & B)\dfrac{4}{3}\lambda \\
 & C)4\lambda \\
 & D)8\lambda \\
\end{align}$

Answer 1

Hint: We will use Einstein’s photoelectric equation to solve this question. We need to make two expressions for the two conditions given. Then by equating those two equations, we can find the threshold wavelength for the given surface. We must know that work function is inversely related to threshold wavelength.

Formula used:
$K.E=h\upsilon -h{{\upsilon }_{0}}$

Complete step-by-step solution:
According to Einstein’s photoelectric equation, the energy of a photon will be the total energy needed to remove an electron from the metallic surface, and the kinetic energy of the emitted electron. But, we must know that stopping potential will be equal to this kinetic energy possessed by the electron because that is the energy we needed to apply for ceasing the motion of that electron.
Then, we know that frequency is inversely related to wavelength and given by the equation, $\upsilon =\dfrac{c}{\lambda }$. Where $c$ is the velocity of light in a vacuum.
Now, we can write the photoelectric equation for the first case as,
$3{{V}_{0}}=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}}$ --- (1)
Where, $3{{V}_{0}}$ is the stopping potential
$h$ is the Planck's constant given as $6.626\times {{10}^{-34}}$ Js
$\lambda $ is the wavelength of the illuminated light
${{\lambda }_{0}}$ is the threshold frequency.
Similarly, for the second case, the illuminated wavelength becomes $2\lambda $ and stopping potential becomes ${{V}_{0}}$. Then, photoelectric equation is given as,
${{V}_{0}}=\dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{0}}}$ --- (2)
Now, we will subtract (1) from (2) multiplied 3 times,
\[\begin{align}
  & \Rightarrow \left( 3{{V}_{0}}-3{{V}_{0}} \right)=3\left( \dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{0}}} \right)-\left( \dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}} \right) \\
 & \Rightarrow \left( \dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}} \right)=3\left( \dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{0}}} \right) \\
 & \Rightarrow \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}}=\dfrac{3}{2}.\dfrac{1}{\lambda }-3\dfrac{1}{{{\lambda }_{0}}} \\
 & \Rightarrow \dfrac{2}{{{\lambda }_{0}}}=\dfrac{1}{2\lambda } \\
 & \Rightarrow {{\lambda }_{0}}=4\lambda \\
\end{align}\]
Therefore, the threshold wavelength of the surface is found to be \[4\lambda \]. Hence, option c is the right choice.

Note: While subtracting between the two equations, we must not cancel the term containing threshold frequency. We must try to eliminate the stopping potential. That’s why we are making the coefficient of stopping potential in both expressions equal. Also, it is not necessary to make the conversion of the work function into threshold frequency at the beginning of the solution. We can find the expression for the work function from the two equations and convert it to threshold frequency at the last.