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: Understanding the Question:
We have three wavelengths: \(\lambda_1\) = series limit of Lyman (transition from \(n=\infty\) to \(n=1\)), \(\lambda_2\) = first line of Lyman (transition from \(n=2\) to \(n=1\)), \(\lambda_3\) = series limit of Balmer (transition from \(n=\infty\) to \(n=2\)). We need the relation.

Step 2: Key Formula or Approach:
For hydrogen atom, \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\).
Thus: \[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H, \] \[ \frac{1}{\lambda_2} = R_H \left( 1 - \frac{1}{4} \right) = \frac{3R_H}{4}, \] \[ \frac{1}{\lambda_3} = R_H \left( \frac{1}{4} - 0 \right) = \frac{R_H}{4}. \] Observe: \(\frac{1}{\lambda_2} - \frac{1}{\lambda_3} = \frac{3R_H}{4} - \frac{R_H}{4} = \frac{2R_H}{4} = \frac{R_H}{2}\)? That does not match \(R_H\). Let's check carefully. Actually \(\lambda_1\) series limit Lyman = \(n_1=1, n_2=\infty\) gives \(1/\lambda_1 = R_H\). \(\lambda_2\) first line Lyman: \(n_1=1, n_2=2\) gives \(1/\lambda_2 = R_H(1 - 1/4) = 3R_H/4\). \(\lambda_3\) series limit Balmer: \(n_1=2, n_2=\infty\) gives \(1/\lambda_3 = R_H(1/4 - 0) = R_H/4\). Then note: \(1/\lambda_1 = 1/\lambda_2 + 1/\lambda_3\)? Check: \(3R_H/4 + R_H/4 = R_H\). Yes! So \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}\). Rearranging: \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}\) ⇒ \(\frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{1}{\lambda_3}\)? Wait, that gives \( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} = \frac{1}{\lambda_3}\)? But from above: \(1/\lambda_2 = 3R_H/4\), \(1/\lambda_1 = R_H\), difference = \(-R_H/4\) negative. So that's not correct. Let's find which option matches: \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} - \frac{1}{\lambda_3}\) gives RHS = \(3R_H/4 - R_H/4 = R_H/2\) not equal to \(R_H\). So not (A). Option (B): \(\frac{1}{\lambda_1} = \frac{1}{\lambda_2} - \frac{1}{\lambda_3}\)? That gave \(R_H = R_H/2\) false. Wait recalc: Actually \(1/\lambda_2 = 3R_H/4\), \(1/\lambda_3 = R_H/4\), so \(1/\lambda_2 - 1/\lambda_3 = (3R_H/4 - R_H/4) = R_H/2\). Not equal to \(1/\lambda_1 = R_H\). So (B) false. Option (A): \(\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}\) gives \(R_H - 3R_H/4 = R_H/4 = 1/\lambda_3\) correct! Yes. So \(\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}\). That is option (A). But let's double-check: \(1/\lambda_1 = R_H\), \(1/\lambda_2 = 3R_H/4\), difference = \(R_H/4\) = \(1/\lambda_3\). So (A) is correct. Earlier I mistakenly wrote (B). The correct relation is \(\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}\).

Step 3: Detailed Explanation:
Substituting values verifies option (A).

Step 4: Final Answer:
Option (A) is correct.