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COMEDK · Physics · 27. Dual Nature of Matter

Threshold wavelength for photoelectric emission from a metal surface is \(5200\) \(Å\). Photoelectrons will be emitted, when this surface is illuminated with monochromatic radiation from

  1. A \(1 \mathrm{~W~IR}\) lamp
  2. B \(50 \mathrm{~W~UV}\) lamp
  3. C \(50 \mathrm{~W~IR}\) lamp
  4. D \(10 \mathrm{~W~IR}\) lamp
Verified Solution

Answer & Solution

Correct Answer

(B) \(50 \mathrm{~W~UV}\) lamp

Step-by-step Solution

Detailed explanation

Given, \(\lambda_{0}=5200 Å=5200 \times 10^{-10} \mathrm{~m}\) \(\therefore\) Threshold frequency, \(f_{0}=\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8}}{5200 \times 10^{-10}}\) \(=5.76 \times 10^{14} \mathrm{~Hz}\) It lies in the frequency range of UV-rays, so…