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Physics Question on Bohr’s Model for Hydrogen Atom

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=1.6×10−19 Chc = 1245 \, \text{eV nm}, \, e = 1.6 \times 10^{-19} \, \text{C}).

Answer

The energy corresponding to the first dip is given as:
10.2 eV=hcλ10.2 \, \text{eV} = \frac{hc}{\lambda}
Rearrange to solve for λ\lambda:
λ=1245 eV⋅nm10.2 eV\lambda = \frac{1245 \, \text{eV} \cdot \text{nm}}{10.2 \, \text{eV}}
λ=122.06 nm\lambda = 122.06 \, \text{nm}