In photoelectric experiment the stopping potential was measured to be $V_1$ and $V_2$ volt with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively the value of $V_2$ is [ $(\phi=$ work function, $e=$ electronic charge $]$

Question

Aicharya · Accepted Answer

Einstein's' photoelectric equation dictates:
$e V=\frac{h c}{\lambda}-\phi$
For incident wavelength $\lambda$ :
$\frac{h c}{\lambda}=\phi+e V_1 \quad---(1)$
For incident wavelength $\frac{\lambda}{2}$ :
$\frac{h c}{\lambda / 2}=\phi+e V_2$
$2 \frac{h c}{\lambda}=\phi+e V_2 \quad--(2)$
On subtracting equation (2) and (1)
$\begin{aligned} & \frac{2}{1}=\frac{\phi+e V_2}{\phi+e V_1} \\ & \Rightarrow 2 \phi+2 e V_1=\phi-e V_2 \\ & \Rightarrow V_2=\frac{\phi}{e}+2 V_1\end{aligned}$ ~

In photoelectric experiment the stopping potential was measured to be $V_1$ and $V_2$ volt with incident…

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