MHT CET · Physics · Dual Nature of Matter
In the case of photoelectric emission from a certain metal, the cutoff frequency is \(v\). If radiation of frequency \(2 v\) is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ( \(m=\) mass of electron)
- A \(2 \sqrt{\frac{h v}{m}}\)
- B \(\sqrt{\frac{h v}{2 m}}\)
- C \(\sqrt{\frac{2 h v}{m}}\)
- D \(\sqrt{\frac{h v}{m}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{2 h v}{m}}\)
Step-by-step Solution
Detailed explanation
\(K \cdot E .=E-\phi\)
\(\frac{1}{2} m v_{\max }^2=h(2 v)-h v\)
\(v_{\max }=\sqrt{\frac{2 h v}{m}}\)
\(\frac{1}{2} m v_{\max }^2=h(2 v)-h v\)
\(v_{\max }=\sqrt{\frac{2 h v}{m}}\)
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