Question 312: Dual Nature of Matter and Radiation

Question

When light of wavelength $$\lambda$$ is incident on a photosensitive surface the stopping potential is '$$\mathrm{V}$$'. When light of wavelength $$3 \lambda$$ is incident on same surface the stopping potential is $$\frac{\mathrm{V}}{6}$$. Then the threshold wavelength for the surface is

Accepted Answer

$$\begin{aligned}
& \lambda_1=\lambda,\left(\mathrm{V}_0ight)_1=\mathrm{V} \
& \lambda_2=3 \lambda,\left(\mathrm{V}_0ight)_2=\frac{\mathrm{V}}{6}
\end{aligned}$$
Photo electric equation is given by
$$\mathrm{eV}_0=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight)$$ ..... (i)
In first case,
$$\mathrm{eV}=\mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_0}ight)$$
For second case,
$$\frac{\mathrm{eV}}{6}=\mathrm{hc}\left(\frac{1}{3 \lambda}-\frac{1}{\lambda_0}ight)$$ ..... (ii)
Dividing equation (i) by equation (ii),
$$\begin{aligned}
6 & =\frac{\frac{1}{\lambda}-\frac{1}{\lambda_0}}{\frac{1}{3 \lambda}-\frac{1}{\lambda_0}}=\frac{3 \lambda_0-\lambda}{\lambda_0-3 \lambda} \
	herefore \quad \lambda_0 & =5 \lambda
\end{aligned}$$

Answer

$$2 \lambda$$

Answer

$$3 \lambda$$

Answer

$$4 \lambda$$

Answer

$$5 \lambda$$

Dual Nature of Matter and Radiation

When light of wavelength $\lambda is incident on a photosensitive surface the stopping potential is '\mathrm{V}'. When light of wavelength 3 \lambda is incident on same surface the stopping potential is \frac{\mathrm{V}}{6}$. Then the threshold wavelength for the surface is

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