Question

In hydrogen spectrum, the shortest wavelength in the Balmer series is \( \lambda \). The shortest wavelength in the Bracket series is:

• A. \( 4 \lambda \)
• B. \( 9 \lambda \)
• C. \( 16 \lambda \)
• D. \( 2 \lambda \)

Hint:

Key points to remember:
1. Shortest wavelength = highest energy transition (\( n_i \to \infty \))
2. Balmer series has \( n_f = 2 \), Brackett has \( n_f = 4 \)
3. Wavelength ratios depend on inverse squares of final levels

Answer 1

The Correct Option is A

Step 1: Understand spectral series
For hydrogen atom transitions, wavelength is given by: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \] where \( R_H \) is Rydberg constant, \( n_f \) is final level, and \( n_i \) is initial level.
Step 2: Shortest wavelength in Balmer series
Shortest wavelength corresponds to maximum energy transition (\( n_i \to \infty \) to \( n_f = 2 \)): \[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty} \right) = \frac{R_H}{4} \] \[ \Rightarrow \lambda = \frac{4}{R_H} \]
Step 3: Shortest wavelength in Brackett series
For Brackett series (\( n_f = 4 \)), shortest wavelength occurs when \( n_i \to \infty \): \[ \frac{1}{\lambda_B} = R_H \left( \frac{1}{4^2} - \frac{1}{\infty} \right) = \frac{R_H}{16} \] \[ \Rightarrow \lambda_B = \frac{16}{R_H} \]
Step 4: Find the ratio \[ \frac{\lambda_B}{\lambda} = \frac{16/R_H}{4/R_H} = 4 \] \[ \Rightarrow \lambda_B = 4\lambda \]