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JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter

The difference between threshold wavelengths for two metal surfaces \(A\) and \(B\) having work function \(\phi_A=9\,eV\) and \(\phi_{ B }=4.5\,eV\) in \(nm\) is:(Given, \(hc =1242\,eV\,nm\))

  1. A \(264\)
  2. B \(138\)
  3. C \(276\)
  4. D \(540\)
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Answer & Solution

Correct Answer

(B) \(138\)

Step-by-step Solution

Detailed explanation

\(\lambda_{ A }=\left(\frac{1242}{9}\right)=138\,nm\) \(\lambda_{ B }=\left(\frac{1242}{4.5}\right)=276\,nm\) \(\lambda_{ B }-\lambda_{ A }=138\,nm\)
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