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

We need to find the wavelength of light emitted when hydrogen is excited to the first excitation level (10.2 eV).

In the Franck-Hertz experiment, the first dip at 10.2 V corresponds to the energy required to excite hydrogen from the ground state $$n = 1$$ to the first excited state $$n = 2$$, and when the atom de-excites back to $$n = 1$$, it emits a photon with energy equal to 10.2 eV.

The energy-wavelength relationship for a photon is given by

$$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$$

First, the energy of the emitted photon is $$E = 10.2 \, \text{eV}$$.

Next, using the value $$hc = 1245 \, \text{eV·nm}$$, we find

$$\lambda = \frac{1245}{10.2} = 122.06 \approx 122 \, \text{nm}$$

This corresponds to the Lyman-alpha line of hydrogen (the $$n = 2 \to n = 1$$ transition), which lies in the ultraviolet region.

The correct answer is 122 nm.