Question

The ratio of the wavelengths of the spectral lines in the Lyman series of the hydrogen spectrum when the transitions take place from 7^th and 9^th states to the ground state is:

• A. 245 : 243
• B. 216 : 293
• C. 251 : 236
• D. 247 : 224

Hint:

For spectral lines, use the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Wavelengths are inversely proportional to $\frac{1}{\lambda}$.

Answer 1

The Correct Option is A

For the Lyman series, transitions are to the ground state ($n_1 = 1$). 
Wavelength is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, where $R$ is the Rydberg constant. 
For $n_2 = 7$: $\frac{1}{\lambda_1} = R \left( 1 - \frac{1}{7^2} \right) = R \left( 1 - \frac{1}{49} \right) = R \cdot \frac{48}{49}$. 
For $n_2 = 9$: $\frac{1}{\lambda_2} = R \left( 1 - \frac{1}{9^2} \right) = R \left( 1 - \frac{1}{81} \right) = R \cdot \frac{80}{81}$. 
Since $\lambda \propto \frac{1}{\frac{1}{\lambda}}$, ratio $\lambda_1 : \lambda_2 = \frac{49}{48} : \frac{81}{80}$. 
Convert to whole numbers: $\lambda_1 : \lambda_2 = \frac{49}{48} \times \frac{80}{81} : 1 \times \frac{80}{81} = \frac{49 \times 80}{48 \times 81} : \frac{80}{81} = \frac{49 \times 80}{48 \times 81} : 1$. 
Calculate: $49 \times 80 = 3920$, $48 \times 81 = 3888$, so $\frac{3920}{3888} = \frac{245}{243}$. 
Thus, $\lambda_1 : \lambda_2 = 245 : 243$.