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

Step 1: Use the hydrogen-like formula.
For one-electron species the wavenumber is \[ \frac{1}{\lambda} = R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right). \]
Step 2: Fix the transition.
Both are first lines of the Lyman series, so $n_1 = 1$, $n_2 = 2$ for both. The bracket term is identical for the two species.

Step 3: Simplify.
Since the bracket and $R$ are the same, \[ \frac{1}{\lambda} \propto Z^2, \] which means \[ \lambda \propto \frac{1}{Z^2}. \]
Step 4: Read off the charges.
For $Li^{2+}$, $Z = 3$. For deuterium $^2_1H$, $Z = 1$.

Step 5: Take the ratio.
\[ \frac{\lambda_{Li^{2+}}}{\lambda_D} = \frac{Z_D^2}{Z_{Li}^2} = \frac{1^2}{3^2} = \frac{1}{9}. \]
Step 6: State the ratio.
So $\lambda_{Li^{2+}} : \lambda_D = 1 : 9$.
\[ \boxed{1 : 9} \]