Physics · Atomic Physics

JEE Main 2024 — 6 April, Shift 2 — Question 51

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 $\qquad$ nm .

(Given hc $=1245 \mathrm{eV} \mathrm{nm}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ ).

Answer: 122

Numerical answer — enter this value.

$10.2 \mathrm{eV}=\frac{\mathrm{hc}}{\lambda}$

$\lambda=\frac{1245 \mathrm{eV}-\mathrm{nm}}{10.2 \mathrm{eV}}=122.06 \mathrm{~nm}$

Answer key and solution verified before publishing.

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