MHT CET · Physics · Dual Nature of Matter
When a light of wavelength \(\lambda\) falls on the emitter of a photocell, maximum speed of emitted photoelectrons is V. If the incident wavelength is changed to \(2 \lambda / 3\), maximum speed of emitted photoelectrons will be
- A \(\sqrt{3} \mathrm{~V}\)
- B \(\frac{V}{2}\)
- C V
- D \(\sqrt{\frac{3}{2}} \mathrm{~V}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{3}{2}} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\( \frac{hc}{\lambda} = \frac{1}{2}mV^2 \) \( \frac{hc}{2\lambda/3} = \frac{1}{2}mV'^2 \)
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