MHT CET · Physics · Dual Nature of Matter
The light of wavelength ' \(\lambda\) ' is incident on the surface of metal of work function \(\phi\) and emits the electron, The maximum velocity of electron emitted is [ \(\mathrm{m}=\) mass of electron and \(\mathrm{h}=\) Planck's constant, \(c=\) velocity of light]
- A \(\left[\frac{2(\mathrm{hc}-\lambda)^{\frac{1}{2}}}{\mathrm{~m} \lambda}\right]\)
- B \(\left[\frac{2(\mathrm{hc}-\phi) \lambda}{\mathrm{mc}}\right]\)
- C \(\left[\frac{2(\mathrm{hc}-\lambda)}{\mathrm{m} \lambda}\right]\)
- D \(\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(D) \(\left[\frac{2(\mathrm{hc}-\lambda \phi)}{\mathrm{m} \lambda}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
We have \(\frac{1}{2} \mathrm{mv}_{\max }^2=\frac{\mathrm{hc}}{\lambda}-\phi=\frac{\mathrm{hc}-\phi \lambda}{\lambda}\)
\(
\begin{aligned}
& \therefore \mathrm{V}_{\max }^2=\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda} \\
& \therefore \mathrm{V}_{\max }=\sqrt{\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda}}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \mathrm{V}_{\max }^2=\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda} \\
& \therefore \mathrm{V}_{\max }=\sqrt{\frac{2(\mathrm{hc}-\phi \lambda)}{\mathrm{m} \lambda}}
\end{aligned}
\)
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