Number of Spectral Emission Lines for Irradiated Hydrogen Atom

To find the total number of spectral emission lines, we first determine the principal quantum number ($n$) of the excited energy state reached by the hydrogen atom after absorbing the incident light.

1. Calculate the energy of the absorbed photon: The energy of a photon of wavelength $\lambda$ is given by: $E = \frac{hc}{\lambda}$

Given:

• $\frac{hc}{e} = 1.237 \times 10^{-6} \text{ eV m}$
• $\lambda = 970 \text{ \AA} = 970 \times 10^{-10} \text{ m}$

Substituting the given values: $E = \frac{1.237 \times 10^{-6} \text{ eV m}}{970 \times 10^{-10} \text{ m}} = \frac{12370}{970} \text{ eV} \approx 12.75 \text{ eV}$

2. Determine the principal quantum number ($n$) of the excited state: The energy of the electron in the ground state ($n_1 = 1$) is: $E_1 = -13.6 \text{ eV}$

The energy of the electron in the excited state $n$ is: $E_n = E_1 + E$ $E_n = -13.6 \text{ eV} + 12.75 \text{ eV} = -0.85 \text{ eV}$

Using Bohr's energy formula for the hydrogen atom: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Equating the two expressions for $E_n$: $-\frac{13.6}{n^2} = -0.85$ $n^2 = \frac{13.6}{0.85} = 16$ $n = 4$

3. Calculate the number of spectral lines: The total number of spectral emission lines $N$ possible during the de-excitation from energy level $n$ to the ground state is given by the formula: $N = \frac{n(n - 1)}{2}$

For $n = 4$: $N = \frac{4 \times (4 - 1)}{2} = \frac{4 \times 3}{2} = 6$

Thus, the total number of lines present in the emission spectrum is 6.