JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Modern Physics

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In Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at 10.2 V. The wavelength of light emitted by hydrogen atom when excited to the first excitation level is ______ nm. (Given hc

= 1245 eV nm, e = 1.6 \times 1 0^{- 19} C

Answer: 122

Solution

The first dip (at 10.2 V) is the first excitation energy of hydrogen (

n = 1 \to n = 2

), i.e.

E = 10.2

eV. The emitted wavelength is

λ = \frac{h c}{E} = \frac{1245}{10.2} \approx 122.06 nm \approx 122 nm

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer

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