Question

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

Answer 1

Correct Answer: 122

The energy corresponding to the first dip is given as:
\[10.2 \, \text{eV} = \frac{hc}{\lambda}\]
Rearrange to solve for \( \lambda \):
\[\lambda = \frac{1245 \, \text{eV} \cdot \text{nm}}{10.2 \, \text{eV}}\]
\[\lambda = 122.06 \, \text{nm}\]