Question

In a photoelectric effect measurement, the stopping potential for a given metal is found to be $V_0$ volt when radiation of wavelength ${λ}_0$ is used. If radiation of wavelength $2{λ}_0$ is used with the same metal then the stopping potential (in volt ) will be

• A. $\frac{V_0}{2}$
• B. $2V_0$
• C. $V_0+\frac{hc}{2e{λ}_0}$
• D. $V_0−\frac{hc}{2e{λ}_0}$

Answer 1

Answer: (D)

When voltage equals stopping potential, kinetic energy of ejected electron is equal to the potential energy at the collector.
According to photoelectric effect,
$e{V}_0=\frac{hc}{{\left(λ\right)}_0}−W_0...\left(1\right)$
and $e{V}^{′}=\frac{hc}{2{\left(λ\right)}_0}−W_0...\left(2\right)$
where, $V^{′}$ is the stopping potential when the metal is irradiated with radiation of wavelength $2{λ}_0$ and $W_0$ is the work function of metal.
Subtracting equation $\left(2\right)$ from $\left(1\right)$ , we get,
$e(V_0−V^{{}^{′}})=\frac{hc}{{\left(λ\right)}_0}[1−\frac{1}{2}]=\frac{hc}{2{\left(λ\right)}_0}$
$⇒V^{{}^{′}}=V_0−\frac{hc}{2e{λ}_0}$