Q 12-12-048JEE MainJEE Main 2024 (6 Apr, Shift 2)Easy
In the Franck-Hertz experiment, the first dip in the current-voltage graph for hydrogen is observed at $10.2\ \text{V}$. The wavelength of light emitted by the hydrogen atom when excited to the first excitation level is ______ nm. (Given $hc = 1245\ \text{eV nm}$, $e = 1.6\times10^{-19}\ \text{C}$)
Numerical value type. Enter your answer.
Answer: 122
The first excitation energy is $10.2\ \text{eV}$, so the emitted photon has
$$\lambda = \frac{1245}{10.2} \approx 122\ \text{nm}$$
Solution by Sreeraj P, M.Sc Physics