Question

The stopping potentials are ${V_1}$ and ${V_2}$ with incident lights of wavelength ${\lambda _1}$ and ${\lambda _2}$ respectively. Then${V_1} - {V_2}$:

A) $\left( {\dfrac{{hc}}{e}} \right) \cdot \left( {\dfrac{{{\lambda _1} \cdot {\lambda _2}}}{{{\lambda _1} - {\lambda _2}}}} \right)$.

B) $\left( {\dfrac{{hc}}{e}} \right) \cdot \left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right)$.

C) $\left( {\dfrac{{he}}{c}} \right) \cdot \left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right)$.

D) $\left( {\dfrac{{he}}{{c{\lambda _1}{\lambda _2}}}} \right) \cdot \left( {{\lambda _1} - {\lambda _2}} \right)$.

Answer 1

Wavelength is the distance between the two consecutive crests or troughs in the wave. The stopping voltage is that voltage required in order to stop the flow of electrons from moving from one plate to another and create current in the photoelectric experiment.

Formula used: The formula of the stopping voltage is given by,

${\text{V}} = \dfrac{{\left( {\dfrac{{hc}}{\lambda } - \phi } \right)}}{e}$

Where stopping voltage is V, Planck’s constant is h, the speed of light is c , the wavelength is $\lambda$, the charge on the electron is $e$ and the work function is $\phi$.

Complete step by step answer:

It is given in the problem that the stopping voltages of the incident light is ${V_1}$ and ${V_2}$ with wavelengths ${\lambda _1}$ and ${\lambda _2}$, we need to find the value of difference of the stopping voltages${V_1} - {V_2}$.

The formula of the stopping voltage is given by,

${\text{V}} = \dfrac{{\left( {\dfrac{{hc}}{\lambda } - \phi } \right)}}{e}$

Where stopping voltage is V, Planck’s constant is h, the speed of light is c, the wavelength is $\lambda$, the charge on the electron is $e$ and the work function is $\phi$.

The stopping voltage corresponding to incident light of wavelength ${\lambda _1}$ is equal to,

${V_1} = \dfrac{{\left( {\dfrac{{hc}}{{{\lambda _1}}} - \phi } \right)}}{e}$………eq. (1)

The stopping voltage corresponding to incident light of wavelength ${\lambda _2}$ is equal to,

${V_2} = \dfrac{{\left( {\dfrac{{hc}}{{{\lambda _2}}} - \phi } \right)}}{e}$………eq. (2)

Taking difference of the stopping voltage we get,

$\Rightarrow {V_1} - {V_2}$

Replacing the value of the stopping voltage from equation (1) and equation (2) in the above relation we get,

$\Rightarrow {V_1} - {V_2}$

$\Rightarrow \left[ {\dfrac{{\left( {\dfrac{{hc}}{{{\lambda _1}}} - \phi } \right)}}{e}} \right] - \left[ {\dfrac{{\left( {\dfrac{{hc}}{{{\lambda _2}}} - \phi } \right)}}{e}} \right]$

$\Rightarrow \dfrac{{hoc}}{{ecdot {\lambda _1}}} - \dfrac{\phi }{e} - \dfrac{{hoc}}{{ecdot {\lambda _ 2}}} + \dfrac{\phi }{e}$

$\Rightarrow \dfrac{{hc}}{{e \cdot {\lambda _1}}} - \dfrac{{hc}}{{e \cdot {\lambda _2}}}$

$\Rightarrow \dfrac{{hc}}{e}\left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right)$.

The difference in the stopping potential is equal to$\dfrac{{hc}}{e}\left( {\dfrac{1}{{{\lambda _1}}} - \dfrac{1}{{{\lambda _2}}}} \right)$.

The correct answer for this problem is option B.

Note: It is advisable for students to understand and remember the formula of the stopping voltage as it is helpful in solving problems like these. Few physical quantities are constant in this formula of stopping voltage like Planck’s constant h and the value of charge of electron e. Also the value of the speed of light will also be constant. So it is also advisable to remember these values because sometimes in the problem, it is not provided in the problem.