Question 324: Dual Nature of Matter and Radiation

Question

When radiation of wavelength '$$\lambda$$' is incident on a metallic surface, the stopping potential is 4.8 V. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes $$1.6 \mathrm{~V}$$. The threshold wavelength for the surface is

Accepted Answer

Stopping potential:
$$\begin{aligned}
& \mathrm{eV}_0=\mathrm{h} 
u-\phi_0 \
& \mathrm{eV}_0=\frac{\mathrm{hc}}{\lambda}-\phi_0
\end{aligned}$$
For first case: $$e(4.8)=\frac{h c}{\lambda}-\phi_0$$ ..... (i)
For Second case: $$e(1.6)=\frac{h c}{2 \lambda}-\phi_0$$ ..... (ii)
Dividing equation (i) by equation (ii),
$$3\left(\frac{\mathrm{hc}}{2 \lambda}-\phi_0ight)=\frac{\mathrm{hc}}{\lambda}-\phi_0$$
$$\begin{aligned}
	herefore \quad & \frac{3 \mathrm{hc}}{2 \lambda}-3 \phi_0=\frac{\mathrm{hc}}{\lambda}-\phi_0 \
& \frac{\mathrm{hc}}{2 \lambda}=2 \phi_0 \
& \frac{\mathrm{hc}}{4 \lambda}=\phi_0
\end{aligned}$$
$$	herefore \quad$$ Threshold wavelength is $$4 \lambda$$.

Answer

$$2 \lambda$$

Answer

$$4 \lambda$$

Answer

$$6 \lambda$$

Answer

$$8 \lambda$$

Dual Nature of Matter and Radiation

When radiation of wavelength '$\lambda' is incident on a metallic surface, the stopping potential is 4.8 V. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes 1.6 \mathrm{~V}$. The threshold wavelength for the surface is

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