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JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
  • JEE Mains 2019
  • Medium
  • MCQ

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 ............. \(nm\)

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Correct Answer

(D) \(250\)

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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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