Question

The ratio of the shortest wavelengths of Brackett and Balmer series of hydrogen atom is

• A. \( 2 : 1 \)
• B. \( 3 : 2 \)
• C. \( 4 : 1 \)
• D. \( 6 : 5 \)

Hint:

To compare wavelengths of hydrogen spectral lines, use \( \lambda \propto \left(\frac{1}{R \left(1 - \frac{1}{n^2}\right)}\right) \). Shortest wavelength means \( n_2 \to \infty \).

Answer 1

The Correct Option is C

Step 1: Use the formula for shortest wavelength in any series: \[ \frac{1}{\lambda_{\text{min}}} = R \left(1 - \frac{1}{n^2}\right) \] where \( n \) is the principal quantum number of the lower energy level of the series. Step 2: For Balmer series, \( n = 2 \) \[ \frac{1}{\lambda_B} = R \left(1 - \frac{1}{2^2}\right) = R \left(\frac{3}{4}\right) \] Step 3: For Brackett series, \( n = 4 \) \[ \frac{1}{\lambda_{Br}} = R \left(1 - \frac{1}{4^2}\right) = R \left(\frac{15}{16}\right) \] Step 4: Take ratio of wavelengths (remember it's inverse of above) \[ \frac{\lambda_{Br}}{\lambda_B} = \frac{\frac{4}{3}}{\frac{16}{15}} = \frac{4}{3} . \frac{15}{16} = \frac{5}{4} \Rightarrow \lambda_B : \lambda_{Br} = 1 : 4 \Rightarrow \lambda_{Br} : \lambda_B = 4 : 1 \]