Question 336: Dual Nature of Matter and Radiation

Question

The maximum kinetic energies of photoelectrons emitted are $$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ when lights of wavelengths $$\lambda_1$$ and $$\lambda_2$$ are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$ then

Accepted Answer

Kinetic energy of the photoelectrons
$$\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi$$
$$	herefore \quad$$ For wavelength $$\lambda_1$$,
$$\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\phi$$ ..... (i)
$$	herefore \quad$$ For wavelength $$\lambda_1$$,
$$\mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\phi$$ ..... (ii)
Subtracting equation (i) from equation (ii),
$$\begin{aligned}
& \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\phi-\frac{\mathrm{hc}}{\lambda_1}+\phi \
& \mathrm{K}_2-\mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_2}-\frac{\mathrm{hc}}{3 \lambda_2} \
& \mathrm{~K}_2-\mathrm{K}_1=\frac{2}{3} \frac{\mathrm{hc}}{\lambda_2} \
& \frac{\mathrm{hc}}{\lambda_2}=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1ight) \quad 	ext{..... (iii)}
\end{aligned}$$
Substituting equation (iii) in equation (ii),
$$\mathrm{K}_2=\frac{3}{2}\left(\mathrm{~K}_2-\mathrm{K}_1ight)-\phi$$
$$\begin{aligned}
& 2 \mathrm{~K}_2=3 \mathrm{~K}_2-3 \mathrm{~K}_1-2 \phi \
& \mathrm{K}_2-3 \mathrm{~K}_1=2 \phi \
& \mathrm{As}, \phi>0 \
& \mathrm{~K}_2-3 \mathrm{~K}_1>0 \
& \mathrm{~K}_1<\frac{\mathrm{K}_2}{3}
\end{aligned}$$

Answer

$$\mathrm{K}_1=\frac{\mathrm{K}_2}{3}$$

Answer

$$\mathrm{K}_1<\frac{\mathrm{K}_2}{3}$$

Answer

$$\mathrm{K}_1=3 \mathrm{K}_2$$

Answer

$$3 \mathrm{K}_1=2 \mathrm{K}_2$$

Dual Nature of Matter and Radiation

The maximum kinetic energies of photoelectrons emitted are $\mathrm{K}_1 and \mathrm{K}_2 when lights of wavelengths \lambda_1 and \lambda_2 are incident on a metallic surface. If \lambda_1=3 \lambda_2$ then

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