Question

Draw energy level diagram for hydrogen atom. Show the transitions of the first line of Lyman series and second line of Balmer series. Find out the ratio of their wavelengths.

Hint:

For hydrogen atom transitions, remember the energy of the transition is given by \( \Delta E = E_{\text{final}} - E_{\text{initial}} \), and the wavelength is related to the energy by \( \lambda = \frac{hc}{\Delta E} \).

Answer 1

In the hydrogen atom, the energy levels are given by the formula: \[ E_n = - \frac{13.6}{n^2} \, \text{eV} \] Where: - \( E_n \) is the energy of the nth level, - \( n \) is the principal quantum number, - 13.6 eV is the Rydberg energy for hydrogen.
1. Energy Level Diagram for Hydrogen:
The energy levels in the hydrogen atom are shown in the diagram below: In this diagram: - The energy of the electron decreases as it moves closer to the nucleus.
- The ground state corresponds to \( n = 1 \), and higher levels correspond to \( n = 2, 3, 4, .... \).
2. Lyman Series Transitions:
The Lyman series corresponds to transitions where the final state is \( n = 1 \) (the ground state). The first line of the Lyman series occurs when the electron transitions from \( n = 2 \) to \( n = 1 \). The energy difference between these levels is: \[ \Delta E_1 = E_2 - E_1 = \left( - \frac{13.6}{2^2} \right) - \left( - \frac{13.6}{1^2} \right) = -3.4 \, \text{eV} + 13.6 \, \text{eV} = 10.2 \, \text{eV} \] The wavelength \( \lambda_1 \) of this transition is given by the formula: \[ \lambda_1 = \frac{hc}{\Delta E_1} \] Substituting the values (\( h = 6.626 \times 10^{-34} \, \text{J} . \text{s} \), \( c = 3 \times 10^8 \, \text{m/s} \), and \( \Delta E_1 = 10.2 \, \text{eV} \)): \[ \lambda_1 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{10.2 \times 1.602 \times 10^{-19}} \approx 121.6 \, \text{nm} \] 3. Balmer Series Transitions:
The Balmer series corresponds to transitions where the final state is \( n = 2 \). The second line of the Balmer series occurs when the electron transitions from \( n = 3 \) to \( n = 2 \). The energy difference is: \[ \Delta E_2 = E_3 - E_2 = \left( - \frac{13.6}{3^2} \right) - \left( - \frac{13.6}{2^2} \right) = -1.511 \, \text{eV} + 3.4 \, \text{eV} = 1.889 \, \text{eV} \] The wavelength \( \lambda_2 \) of this transition is: \[ \lambda_2 = \frac{hc}{\Delta E_2} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.889 \times 1.602 \times 10^{-19}} \approx 656.3 \, \text{nm} \] 4. Ratio of the Wavelengths:
The ratio of the wavelengths is: \[ \frac{\lambda_1}{\lambda_2} = \frac{121.6 \, \text{nm}}{656.3 \, \text{nm}} \approx 0.185 \] Thus, the ratio of the wavelengths of the first line of the Lyman series and the second line of the Balmer series is approximately 0.185.