Question

When a metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. if the same surface is illuminated with radiation of wavelength $2\lambda$ , the stopping potential is $\dfrac{V}{4}$. The threshold wavelength for metallic surface is:
$\begin{aligned} & a)4\lambda \\\ & b)5\lambda \\\ & c)\dfrac{5}{2}\lambda \\\ & d)3\lambda \\\ \end{aligned}$

Answer 1

In the question it is mentioned that a surface is illuminated with different wavelengths of light. The threshold wavelength for the surface remains a constant. Hence we will consider the energy given to the surface by both the wavelengths of light and accordingly obtain the threshold wavelength.
Formula used:
$E=W+e{{V}_{\circ }}$
$E=\dfrac{hc}{\lambda }$
$W=\dfrac{hc}{{{\lambda }_{\circ }}}$

Complete step-by-step solution:
Let us say we illuminate the surface of a metal with a wavelength of light equal to $\lambda$. Then the energy (E) transferred to the surface of the metal from the photons of light is given by,$E=\dfrac{hc}{\lambda }$ where ‘c’ is the speed of light in vacuum and ‘h’ is the Planck’s constant. The metal basically requires some amount of energy in order to ensure the electrons just escape the metal surface i.e. the work function. The threshold wavelength refers to the maximum value of wavelength ${{\lambda }_{\circ }}$ of the incident light required to overcome the work function. Mathematically we can write work function as,
$W=\dfrac{hc}{{{\lambda }_{\circ }}}$
Let us say for wavelength $\lambda$, the stopping potential is ${{V}_{\circ }}$. Therefore using the conservation of energy principle we get,
$E=W+e{{V}_{\circ }}$, where e is the charge on an electron.
The above equation can also be written as,
$\begin{aligned} & E=W+e{{V}_{\circ }} \\\ & \Rightarrow \dfrac{hc}{\lambda }=\dfrac{hc}{{{\lambda }_{\circ }}}+e{{V}_{\circ }}...(1) \\\ \end{aligned}$
In the above question it is given to us that surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is V. Hence using equation 1 we get,
$\begin{aligned} & \dfrac{hc}{\lambda }=\dfrac{hc}{{{\lambda }_{\circ }}}+eV \\\ & \Rightarrow eV=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}}...(2) \\\ \end{aligned}$
Similarly when the surface is illuminated with radiation of wavelength $2\lambda$ , the stopping potential is $\dfrac{V}{4}$Hence using equation 1 we get,
$\begin{aligned} & \dfrac{hc}{2\lambda }=\dfrac{hc}{{{\lambda }_{\circ }}}+e\dfrac{V}{4} \\\ & \Rightarrow \dfrac{eV}{4}=\dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}} \\\ & \Rightarrow eV=4\left[ \dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}} \right]...(3) \\\ \end{aligned}$
Equating equation 2 and 3 we get,
$\begin{aligned} &\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}}=4\left[ \dfrac{hc}{2\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}} \right] \\\ & \Rightarrow \dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{\circ }}}=\dfrac{2hc}{\lambda }-\dfrac{4hc}{{{\lambda }_{\circ }}} \\\ & \Rightarrow \dfrac{hc}{\lambda }-\dfrac{2hc}{\lambda }=-\dfrac{4hc}{{{\lambda }_{\circ }}}+\dfrac{hc}{{{\lambda }_{\circ }}} \\\ & \Rightarrow -\dfrac{hc}{\lambda }=-\dfrac{3hc}{{{\lambda }_{\circ }}} \\\ & \Rightarrow {{\lambda }_{\circ }}=3\lambda \\\ \end{aligned}$
Therefore the correct answer of the above question is option d.

Note: The stopping potential is defined as that value of retarding potential for which the photoelectric current becomes zero. At the stopping potential, no photoelectrons are emitted. The work done on the fastest electron must be equal to its kinetic energy.