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 ............. \(nm\)
- A \(1700\)
- B \(2020\)
- C \(220\)
- D \(250\)
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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