MHT CET · Physics · Dual Nature of Matter
When wavelength of incident radiation on the metal surface is reduced from ' \(\lambda_1\) ' to ' \(\lambda_2\) ' the kinetic energy of emitted photoelectrons is tripled. The work function of metal \([\mathrm{h}=\) Plank 's constant, c = velocity of light]
- A \(\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]\)
- B \(\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]\)
- C \(\operatorname{hc}\left[\frac{3 \lambda_1-\lambda_2}{\lambda_1 \lambda_2}\right]\)
- D \(\operatorname{hc}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{hc}}{2}\left[\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right]\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{K}\) represent kinetic energy
\(
\begin{aligned}
& \therefore \mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\
& \text { and } \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\
& \mathrm{~K}_2=3 \mathrm{~K}_1 \\
& \therefore \frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0=3 \frac{\mathrm{hc}}{\lambda_1}-3 \mathrm{~W}_0 \\
& \therefore 2 \mathrm{~W}_0=\frac{3 \mathrm{hc}}{\lambda_1}-\frac{\mathrm{hc}}{\lambda_2} \\
& \therefore \mathrm{W}_0=\frac{\mathrm{hc}}{2}\left(\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right)
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \mathrm{K}_1=\frac{\mathrm{hc}}{\lambda_1}-\mathrm{W}_0 \\
& \text { and } \mathrm{K}_2=\frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0 \\
& \mathrm{~K}_2=3 \mathrm{~K}_1 \\
& \therefore \frac{\mathrm{hc}}{\lambda_2}-\mathrm{W}_0=3 \frac{\mathrm{hc}}{\lambda_1}-3 \mathrm{~W}_0 \\
& \therefore 2 \mathrm{~W}_0=\frac{3 \mathrm{hc}}{\lambda_1}-\frac{\mathrm{hc}}{\lambda_2} \\
& \therefore \mathrm{W}_0=\frac{\mathrm{hc}}{2}\left(\frac{3 \lambda_2-\lambda_1}{\lambda_1 \lambda_2}\right)
\end{aligned}
\)
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