Question 354: Dual Nature of Matter and Radiation

Question

When certain metal surface is illuminated with a light of wavelength $\lambda$, the stopping potential is $V$, When the same surface is illuminated by light of wavelength $2 \lambda$, the stopping potential is $\left(\frac{V}{3}\right)$. The threshold wavelength for the surface is

Accepted Answer

Given for a metal, wavelength of light used $=\lambda$
Stopping potential $=V$
If $\lambda_0$ be the threshold wavelength, then maximum kinetic energy of emitted electrons
$K_{\max }=h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight)\quad	ext{.... (i)}$
Again, wavelength of used light
$$\lambda^{\prime}=2 \lambda$$
Stopping potential, $V^{\prime}=\frac{V}{3}$
then
 $$K_{\max }=h c\left(\frac{1}{\lambda^{\prime}}-\frac{1}{\lambda_0}ight)$$
$$\begin{array}{ll}
\Rightarrow & e V^{\prime}=h c\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight) \
\Rightarrow & \frac{e V}{3}=h c\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight)\quad	ext{.... (ii)}
\end{array}$$
From Eqs. (i) and (ii), we have
$$\begin{array}{r}
\frac{e V}{\frac{e V}{3}}=\frac{h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight)}{h c\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight)} \
3\left(\frac{1}{2 \lambda}-\frac{1}{\lambda_0}ight)=\frac{1}{\lambda}-\frac{1}{\lambda_0}\
\Rightarrow \lambda_0=4\lambda
\end{array}$$
So, threshold wavelength is 4 times of wavelengths of light.

Answer

$\frac{8 \lambda}{3}$

Answer

$\frac{4 \lambda}{3}$

Answer

$4 \lambda$

Answer

$6 \lambda$

Dual Nature of Matter and Radiation

When certain metal surface is illuminated with a light of wavelength \lambda, the stopping potential is V, When the same surface is illuminated by light of wavelength 2 \lambda, the stopping potential is \left(\frac{V}{3}\right). The threshold wavelength for the surface is

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