Question 303: 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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