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MHT CET · Physics · Dual Nature of Matter

Photoelectrons are emitted from a photosensitive surface for the light of wavelengths \(\lambda_{1}=360 \mathrm{~nm}\) and \(\lambda_{2}=600 \mathrm{~nm} .\) What is the ratio of work functions
for lights of wavelength \({ }^{\prime} \lambda_{1}{ }^{\prime}\) to \(\lambda_{2}{ }^{\prime}\)?

  1. A \(6: 1\)
  2. B \(1: 6\)
  3. C \(5: 3\)
  4. D \(3: 5\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5: 3\)

Step-by-step Solution

Detailed explanation

\(\frac{h c}{\lambda_{1}}-\phi_{1}=\mathrm{KE} \quad \lambda_{1}=360 \mathrm{~nm}=360 \times 10^{-9} \mathrm{~m}\)
\(\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{2}=\mathrm{KE} \quad\lambda_{2}=600 \mathrm{~nm}=600 \times 10^{-9} \mathrm{~m}\)
\(\frac{h c}{\lambda_{1}}=\phi_{1} \quad \quad \phi \propto \frac{1}{\lambda}\)
\(\therefore \frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{600}{360}=\frac{10}{6}=\frac{5}{3}\)
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