MHT CET · Physics · Dual Nature of Matter
Photoelectric emission takes place from a certain metal at threshold frequency \(v\). If the radiation of frequency \(4 v\) is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( \(\mathrm{m}=\) mass of photoelectron, \(\mathrm{h}=\) Planck's constant)
- A \(\sqrt{\frac{6 \mathrm{~h} v}{\mathrm{~m}}}\)
- B \(\sqrt{\frac{3 \mathrm{~h} v}{\mathrm{~m}}}\)
- C \(\sqrt{\frac{\mathrm{h} v}{\mathrm{~m}}}\)
- D \(\sqrt{\frac{5 \mathrm{~h} v}{\mathrm{~m}}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{\frac{6 \mathrm{~h} v}{\mathrm{~m}}}\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{2}m v_{max}^2 = hf - hf_0\) \(\frac{1}{2}m v_{max}^2 = h(4v) - hv\)
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