Question

The wavenumber of the first spectral line of the Lyman series of He\(^+\) ion is \( x \) m\(^{-1}\). What is the wavenumber (in m\(^{-1}\)) of the second spectral line of the Balmer series of Li\(^{2+}\) ion?

• A. \( \frac{9x}{16} \)
• B. \( \frac{16x}{9} \)
• C. \( \frac{8x}{27} \)
• D. \( \frac{27x}{8} \)
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Hint:

For spectral line calculations, use the formula for wavenumber: \[ \tilde{\nu} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right). \]

Answer 1

The Correct Option is A

Step 1: Understanding the Rydberg formula The wavenumber \( \tilde{\nu} \) is given by the Rydberg formula: \[ \tilde{\nu} = RZ^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] where \( Z \) is the atomic number, \( R \) is the Rydberg constant, and \( n_1, n_2 \) are energy levels. Step 2: Wavenumber for He\(^+\) (Lyman series, first line) For He\(^+\), the transition from \( n_2 = 2 \) to \( n_1 = 1 \) is: \[ \tilde{\nu}_{\text{He}^+} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( \frac{3}{4} \right) = 3R \] Given that this wavenumber is \( x \), we set: \[ 3R = x \] Step 3: Wavenumber for Li\(^{2+}\) (Balmer series, second line) For Li\(^{2+}\), the second Balmer line corresponds to \( n_2 = 4 \) to \( n_1 = 2 \): \[ \tilde{\nu}_{\text{Li}^{2+}} = R(3^2) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 9R \left( \frac{4}{16} - \frac{1}{16} \right) = 9R \left( \frac{3}{16} \right) = \frac{27R}{16} \] Step 4: Finding the required ratio Dividing by \( x = 3R \): \[ \frac{\tilde{\nu}_{\text{Li}^{2+}}}{x} = \frac{\frac{27R}{16}}{3R} = \frac{9}{16} \] Thus, \( \tilde{\nu}_{\text{Li}^{2+}} = \frac{9x}{16} \). \bigskip