Question

\(\lambda_1\) is the wavelength of series limit of Lyman series, \(\lambda_2\) is the wavelength of the first line of Lyman series and \(\lambda_3\) is the series limit of the Balmer series. Then the relation between \(\lambda_1\), \(\lambda_2\) and \(\lambda_3\) is

• A. \(\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}\)
• B. \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} - \frac{1}{\lambda_3}\)
• C. \(\lambda_2 = \lambda_1 + \lambda_3\)
• D. \(\lambda_1 = \lambda_2 + \lambda_3\)

Hint:

Remember: For Lyman limit \(1/\lambda_1 = R_H\), first Lyman line \(1/\lambda_2 = 3R_H/4\), Balmer limit \(1/\lambda_3 = R_H/4\). Then \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}\) is true, but rearranged as \(\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}\).

Answer 1

The Correct Option is B

Step 1: Write each wave number.
Using $\tfrac{1}{\lambda} = R_H\left(\tfrac{1}{n_1^2}-\tfrac{1}{n_2^2}\right)$: Lyman limit ($1\to\infty$) gives $\tfrac{1}{\lambda_1} = R_H$. First Lyman line ($1\to2$) gives $\tfrac{1}{\lambda_2} = \tfrac{3R_H}{4}$. Balmer limit ($2\to\infty$) gives $\tfrac{1}{\lambda_3} = \tfrac{R_H}{4}$.

Step 2: Look for a sum.
Notice $\tfrac{3R_H}{4} + \tfrac{R_H}{4} = R_H$, so $\tfrac{1}{\lambda_2} + \tfrac{1}{\lambda_3} = \tfrac{1}{\lambda_1}$.

Step 3: Rearrange.
This is the same as $\tfrac{1}{\lambda_1} - \tfrac{1}{\lambda_2} = \tfrac{1}{\lambda_3}$.

Step 4: State the relation.
\[ \boxed{\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}} \]