CBSE · Class 12 · Physics · AtomsCalculate the shortest and longest wavelengths of the spectral lines in the Lyman series of hydrogen atom. (Given: Rydberg constant $R = 1.097 \times 10^7 \text{ m}^{-1}$). Show all intermediate steps.

To calculate the shortest and longest wavelengths of the spectral lines in the Lyman series of a hydrogen atom, we use the Rydberg formula for hydrogen spectrum:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ \nWhere:

• $\lambda$ is the wavelength of the emitted spectral line.
• $R$ is the Rydberg constant = $1.097 \times 10^7 \text{ m}^{-1}$.
• $n_1$ is the principal quantum number of the lower energy level.
• $n_2$ is the principal quantum number of the higher energy level ($n_2 > n_1$). \nFor the Lyman series, the electrons transition to the ground state, which means the lower orbit is fixed at: $$n_1 = 1$$

Part 1: Calculation of the Shortest Wavelength ($\lambda_{\text{min}}$)

\nThe shortest wavelength corresponds to the maximum energy of the emitted photon. According to the Planck-Einstein relation ($E = \frac{hc}{\lambda}$), maximum energy requires the maximum possible energy difference between orbits. This occurs when an electron falls from the outermost infinite orbit to the ground state: $$n_2 = \infty$$ \nSubstituting $n_1 = 1$ and $n_2 = \infty$ into the Rydberg formula: $$\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$$ \nSince $\frac{1}{\infty} = 0$: $$\frac{1}{\lambda_{\text{min}}} = R (1 - 0) = R$$

$$\lambda_{\text{min}} = \frac{1}{R}$| \nNow, substitute the value of $R$: $$\lambda_{\text{min}} = \frac{1}{1.097 \times 10^7 \text{ m}^{-1}}$$

$$\lambda_{\text{min}} = 0.91157 \times 10^{-7} \text{ m}$$

$$\lambda_{\text{min}} = 9.12 \times 10^{-8} \text{ m} \quad \text{or} \quad 91.2 \text{ nm}$|

Part 2: Calculation of the Longest Wavelength ($\lambda_{\text{max}}$)

\nThe longest wavelength corresponds to the minimum energy of the emitted photon. This occurs when the energy difference between consecutive energy levels is the smallest, which happens when an electron transitions from the immediately adjacent higher orbit to the ground state: $$n_2 = 2$$ \nSubstituting $n_1 = 1$ and $n_2 = 2$ into the Rydberg formula: $$\frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$$

$$\frac{1}{\lambda_{\text{max}}} = R \left( 1 - \frac{1}{4} \right)$|

$$\frac{1}{\lambda_{\text{max}}} = R \left( \frac{3}{4} \right)$|

$$\lambda_{\text{max}} = \frac{4}{3R}$| \nNow, substitute the value of $R$: $$\lambda_{\text{max}} = \frac{4}{3 \times 1.097 \times 10^7 \text{ m}^{-1}}$$

$$\lambda_{\text{max}} = \frac{4}{3.291 \times 10^7 \text{ m}^{-1}}$$

$$\lambda_{\text{max}} = 1.21543 \times 10^{-7} \text{ m}$|

$$\lambda_{\text{max}} = 1.215 \times 10^{-7} \text{ m} \quad \text{or} \quad 121.5 \text{ nm}$|

Conclusion:

• The shortest wavelength in the Lyman series is $91.2 \text{ nm}$ (or $912 \text{ Å}$).-
• The longest wavelength in the Lyman series is $121.5 \text{ nm}$ (or $1215 \text{ Å}$).