Question

One of the lines in the emission spectrum of \(\text{Li}^{2+}\) has the same wavelength as that of the second line of the Balmer series in hydrogen spectrum. The Balmer transition corresponds to \(n=4\rightarrow2\). If the corresponding transition in \(\text{Li}^{2+}\) is \(n=12\rightarrow x\), find the value of \(x\).

• A. \(8\)
• B. \(6\)
• C. \(7\)
• D. \(5\)

Hint:

For hydrogen-like ions: \[ \lambda\propto\frac{1}{Z^2} \] Higher \(Z\) compresses the spectrum.

Answer 1

The Correct Option is B

Step 1: Rydberg formula: \[ \frac{1}{\lambda}=RZ^2\!\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \]
Step 2: For hydrogen Balmer line: \[ \frac{1}{\lambda}=R\!\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3R}{16} \]
Step 3: For \(\text{Li}^{2+}\) (\(Z=3\)): \[ 9R\!\left(\frac{1}{x^2}-\frac{1}{12^2}\right)=\frac{3R}{16} \]
Step 4: Solving: \[ \frac{1}{x^2}=\frac{1}{36}\Rightarrow x=6 \]