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

In a Frank-hertz experiment, an electron of energy \(5.6\,eV\) passes through mercury vapour and emerges with an energy \(0.7\,eV.\) The minimum wavelength of photons emitted by mercury atoms is close to ............. \(nm\)

  1. A \(1700\)
  2. B \(2020\)
  3. C \(220\)
  4. D \(250\)
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Answer & Solution

Correct Answer

(D) \(250\)

Step-by-step Solution

Detailed explanation

\(5.6\,\,{\text{eV}} - 0.7{\text{eV}} = 4.9\,{\text{eV}}\) \( = \frac{{12410\,\,{\text{eV}} - \mathop {\text{A}}\limits^o }}{\lambda }\) \(\lambda = \frac{{12410\,\,{\text{eV}} - \mathop {\text{A}}\limits^o }}{{4.9\,{\text{eV}}}}\) \( \approx 250\,{\text{nm}}\)
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