Question

If \(\lambda_{Li^{2+}}\) and \(\lambda_D\) represent the wavelengths related to first line (shortest line) of Lyman series of line spectrum of \(Li^{2+}\) and \(^2_1H\) respectively, then \(\lambda_{Li^{2+}} : \lambda_D\) is

• A. \(1:4\)
• B. \(1:9\)
• C. \(4:1\)
• D. \(9:1\)

Hint:

For hydrogen-like species: \[ \lambda \propto \frac{1}{Z^2} \] Thus, higher nuclear charge produces shorter wavelengths.

Answer 1

The Correct Option is B

Concept: For hydrogen-like species, \[\begin{aligned} \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \end{aligned}\] For the first line of the Lyman series, \[\begin{aligned} n_1=1,\qquad n_2=2 \end{aligned}\] Hence, \[\begin{aligned} \frac{1}{\lambda} \propto Z^2 \end{aligned}\] Therefore, \[\begin{aligned} \lambda \propto \frac{1}{Z^2} \end{aligned}\]

Step 1: Determine the atomic numbers. \[ \begin{aligned} Li^{2+} &:\quad Z=3 \\ {}^{2}_{1}H\;(\text{Deuterium}) &:\quad Z=1 \end{aligned} \]

Step 2: Use the proportionality relation. \[\begin{aligned} \lambda_{Li^{2+}} : \lambda_D = \frac{1}{3^2} : \frac{1}{1^2} \end{aligned}\] \[\begin{aligned} = \frac{1}{9} : 1 \end{aligned}\]

Step 3: Obtain the ratio. \[\begin{aligned} \boxed{ \lambda_{Li^{2+}} : \lambda_D = 1:9 } \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.