Question

If the minimum wavelength for H-atom in the Lyman series is 'x', then the maximum wavelength of the Balmer series of He\(^+\) ion in terms of 'x' will be:

• A. \( \frac{3}{4} x \)
• B. \( x \)
• C. \( \frac{9x}{5} \)
• D. \( \frac{5x}{9} \)

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are asked to compare the shortest wavelength of the Lyman series for Hydrogen with the longest wavelength of the Balmer series for \(He^+\).
Step 2: Key Formula or Approach:
The Rydberg formula: \(\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\).
- Lyman min (\(n_1=1, n_2=\infty\)).
- Balmer max (\(n_1=2, n_2=3\)).
Step 3: Detailed Explanation:
1. For Hydrogen (\(Z=1\)), Lyman series minimum wavelength (\(\lambda = x\)):
\[ \frac{1}{x} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies R = \frac{1}{x} \]
2. For \(He^+\) (\(Z=2\)), Balmer series maximum wavelength (\(\lambda_{max}\)):
\[ \frac{1}{\lambda_{max}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
\[ \frac{1}{\lambda_{max}} = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} \]
Substitute \(R = \frac{1}{x}\):
\[ \frac{1}{\lambda_{max}} = \frac{5}{9x} \implies \lambda_{max} = \frac{9x}{5} \]
Step 4: Final Answer:
The maximum wavelength of Balmer series of \(He^+\) ion is \(\frac{9x}{5}\).