Q. In a photoelectric effect measurement, the stopping potential for a given metal is found to be $V_{0}$ volt when radiation of wavelength $\lambda _{0}$ is used. If radiation of wavelength $2 \lambda _{0}$ is used with the same metal then the stopping potential (in volt ) will be
NTA AbhyasNTA Abhyas 2022
Solution:
When voltage equals stopping potential, kinetic energy of ejected electron is equal to the potential energy at the collector.
According to photoelectric effect,
$eV_{0}=\frac{h c}{\left(\lambda \right)_{0}}-W_{0}...\left(1\right)$
and $eV'=\frac{h c}{2 \left(\lambda \right)_{0}}-W_{0}...\left(2\right)$
where, $V'$ is the stopping potential when the metal is irradiated with radiation of wavelength $2\lambda _{0}$ and $W_{0}$ is the work function of metal.
Subtracting equation $\left(2\right)$ from $\left(1\right)$ , we get,
$e \left(\right. V_{0} - V^{'} \left.\right) = \frac{h c}{\left(\lambda \right)_{0}} \left[\right. 1 - \frac{1}{2} \left]\right. = \frac{h c}{2 \left(\lambda \right)_{0}}$
$\Rightarrow V^{'}=V_{0}-\frac{h c}{2 e \lambda _{0}}$
According to photoelectric effect,
$eV_{0}=\frac{h c}{\left(\lambda \right)_{0}}-W_{0}...\left(1\right)$
and $eV'=\frac{h c}{2 \left(\lambda \right)_{0}}-W_{0}...\left(2\right)$
where, $V'$ is the stopping potential when the metal is irradiated with radiation of wavelength $2\lambda _{0}$ and $W_{0}$ is the work function of metal.
Subtracting equation $\left(2\right)$ from $\left(1\right)$ , we get,
$e \left(\right. V_{0} - V^{'} \left.\right) = \frac{h c}{\left(\lambda \right)_{0}} \left[\right. 1 - \frac{1}{2} \left]\right. = \frac{h c}{2 \left(\lambda \right)_{0}}$
$\Rightarrow V^{'}=V_{0}-\frac{h c}{2 e \lambda _{0}}$