Question

The wavenumber of the first line (\(n_2 = 3\)) in the Balmer series of hydrogen is \( \overline{\nu}_1 \, \text{cm}^{-1} \). What is the wavenumber (in cm\(^{-1}\)) of the second line (\(n_2 = 4\)) in the Balmer series of He\(^{+}\)? 

• A. \( \frac{5\overline{\nu}_1}{27} \)
• B. \( \frac{27\overline{\nu}_1}{5} \)
• C. \( \frac{27\overline{\nu}_1}{20} \)
• D. \( \frac{20\overline{\nu}_1}{27} \)

Hint:

To compare wavenumbers of spectral lines in hydrogen-like species, use the proportionality \( \overline{\nu} \propto Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \).

Answer 1

The Correct Option is B

For hydrogen-like species, the wavenumber is given by: \[ \overline{\nu} \propto Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Given: - For H (\(Z=1\)), first line: \(n_2 = 3 \to n_1 = 2\), wavenumber = \( \overline{\nu}_1 \) \[ \overline{\nu}_1 \propto 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} \] - For He\(^{+}\) (\(Z=2\)), second line: \(n_2 = 4 \to n_1 = 2\) \[ \overline{\nu}_2 \propto 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4 \left( \frac{1}{4} - \frac{1}{16} \right) = 4 \cdot \frac{3}{16} = \frac{12}{16} = \frac{3}{4} \] Now: \[ \frac{\overline{\nu}_2}{\overline{\nu}_1} = \frac{(3/4)}{(5/36)} = \frac{3}{4} \cdot \frac{36}{5} = \frac{108}{20} = \frac{27}{5} \Rightarrow \overline{\nu}_2 = \frac{27\overline{\nu}_1}{5} \]