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: Name the three wavelengths.
$\lambda_1$ is the Lyman series limit (jump from $n=\infty$ to $n=1$), $\lambda_2$ is the first Lyman line (jump from $n=2$ to $n=1$), and $\lambda_3$ is the Balmer series limit (jump from $n=\infty$ to $n=2$). We need the link between them.

Step 2: Recall the Rydberg formula.
For hydrogen, $\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$.

Step 3: Find the first wave number.
Lyman limit: $\frac{1}{\lambda_1} = R_H\left(1 - 0\right) = R_H$.

Step 4: Find the second wave number.
First Lyman line: $\frac{1}{\lambda_2} = R_H\left(1 - \frac{1}{4}\right) = \frac{3R_H}{4}$.

Step 5: Find the third wave number.
Balmer limit: $\frac{1}{\lambda_3} = R_H\left(\frac{1}{4} - 0\right) = \frac{R_H}{4}$.

Step 6: Spot the relation.
Subtract the second from the first: $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = R_H - \frac{3R_H}{4} = \frac{R_H}{4} = \frac{1}{\lambda_3}$. So the clean relation is below.
\[ \boxed{\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}} \]