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MHT CET · Physics · Dual Nature of Matter

When a light of wavelength \(300 \mathrm{~nm}\) fall on a photoelectric emitter, photo electrons are emitted. For another emitter light of wavelength \(600 \mathrm{~nm}\) is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

  1. A \(1: 2\)
  2. B \(2: 1\)
  3. C \(4: 1\)
  4. D \(1: 4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(2: 1\)

Step-by-step Solution

Detailed explanation

Work function \(\phi_0=\frac{\mathrm{hc}}{\lambda_0}\)
\(
\begin{aligned}
\therefore \phi_0 & \propto \frac{1}{\lambda_0} \\
\frac{\phi_{0_1}}{\phi_{0_2}} & =\frac{\lambda_{0_2}}{\lambda_{0_1}}=\frac{600}{300}=\frac{2}{1}
\end{aligned}
\)