Question

When a metallic surface is illuminated with light of wavelength $\lambda $, the stopping potential is x volt. When the same surface is illuminated by light of wavelength $2\lambda $, the stopping potential is $\dfrac{x}{3}$. Threshold wavelength for the metallic surface is:
A) $\dfrac{{4\lambda }}{3}$.
B) $4\lambda $.
C) $6\lambda $.
D) $\dfrac{{8\lambda }}{3}$.

Answer 1

Hint:Stopping voltage is defined as the potential difference that is applied to stop the moving electron from the late and generating current. Threshold frequency is the minimum frequency that must be given in the wave so as to make the flow of electrons from the plate.

Formula used:The formula of the Einstein equation for photoelectric effect is given by,
$ \Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
Where threshold potential difference is ${V_o}$ the threshold wavelength is ${\lambda _o}$ the wavelength of the wave is $\lambda $ the charge of the electron is e the Planck’s constant is h and the speed of light is c.

Complete step by step solution:
It is given in the problem that when a metallic surface is illuminated with light of wavelength $\lambda $, the stopping potential is x volt and the same surface is illuminated by light of wavelength $2\lambda $, the stopping potential is $\dfrac{x}{3}$ and we need to find the threshold wavelength for the metallic surface.
For the metallic surface with wavelength of $\lambda $.
The formula of the Einstein equation for photoelectric effect is given by,
$ \Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
Where threshold potential difference is ${V_o}$ the threshold wavelength is ${\lambda _o}$ the wavelength of the wave is $\lambda $ the charge of the electron is e the Planck’s constant is h and the speed of light is c.
The stopping voltage is x and the wavelength is $\lambda $,
$ \Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{{x \cdot e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$………eq. (1)
For the wavelength of stopping voltage of $\dfrac{x}{3}$ wavelength $2\lambda $.
The formula of the Einstein equation for photoelectric effect is given by,
$ \Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
Where threshold potential difference is ${V_o}$ the threshold wavelength is ${\lambda _o}$ the wavelength of the wave is $\lambda $ the charge of the electron is e the Planck’s constant is h and the speed of light is c.
$ \Rightarrow \dfrac{{{V_o}e}}{{hc}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{{x \cdot e}}{{3\left( {h \cdot c} \right)}} = \dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}$………eq. (2)
Taking ratio of equation (1) and equation (2).
$ \Rightarrow \dfrac{{\left( {\dfrac{{x \cdot e}}{{hc}}} \right)}}{{\left( {\dfrac{{x \cdot e}}{{3\left( {h \cdot c} \right)}}} \right)}} = \dfrac{{\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)}}{{\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right)}}$
$ \Rightarrow 3 = \dfrac{{\left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)}}{{\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right)}}$
$ \Rightarrow 3\left( {\dfrac{1}{{2\lambda }} - \dfrac{1}{{{\lambda _o}}}} \right) = \left( {\dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}} \right)$
$ \Rightarrow \dfrac{3}{{2\lambda }} - \dfrac{3}{{{\lambda _o}}} = \dfrac{1}{\lambda } - \dfrac{1}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{3}{{2\lambda }} = \dfrac{1}{\lambda } + \dfrac{2}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{3}{{2\lambda }} - \dfrac{1}{\lambda } = \dfrac{2}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{{3 - 2}}{{2\lambda }} = \dfrac{2}{{{\lambda _o}}}$
$ \Rightarrow \dfrac{1}{{2\lambda }} = \dfrac{2}{{{\lambda _o}}}$
$ \Rightarrow {\lambda _o} = 4\lambda $.
The threshold frequency is equal to ${\lambda _o} = 4\lambda $.

The correct answer for this problem is option B.

Note:It is advisable for students to understand and remember the formula of the photoelectric effect as it is helpful in solving problems like these. The threshold frequency is the minimum frequency which is required to make the electrons flow.