Question

In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the bracket series is :

• A. 16λ
• B. 2λ
• C. 4λ
• D. 9λ

Answer 1

The Correct Option is C

To determine the shortest wavelength in the Brackett series compared to the Balmer series, we refer to the Rydberg formula for wavelength of emitted light: 

\[\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\]

where \( R \) is the Rydberg constant, \( n_1 \) and \( n_2 \) are integers with \( n_2 > n_1 \). For the Balmer series, the transitions end at \( n_1 = 2 \), and for the Brackett series, transitions end at \( n_1 = 4 \).

For the Balmer series shortest wavelength (transition from infinity to n=2), we have:

\[\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}\]

For the Brackett series shortest wavelength (transition from infinity to n=4), we have:

\[\frac{1}{\lambda_{Brackett}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = \frac{R}{16}\]

Comparing these:

\[\frac{\lambda_{Balmer}}{\lambda_{Brackett}} = \frac{R/4}{R/16} = \frac{16}{4} = 4\]

Thus, the shortest wavelength in the Brackett series is \( 4\lambda \).

Answer 2

The shortest wavelength in the Balmer series occurs when an electron transitions from \(\infin\ \text{to}\ n = 2.\)
\(\because\frac{1}{λ}=Rz^2[\frac{1}{2^2}-\frac{1}{\infin^2}]\)
\(\frac{1}{λ}=\frac{R}{4}\ \ ....(i)\)
The shortest wavelength in the Brackett series occurs when an electron transitions from \(\infty\ \text{to}\ n = 4\).
\(\frac{1}{\lambda}=R(1)^2[\frac{1}{4^2}-\frac{1}{∞^2}]\)
\(⇒\frac{1}{λ'}=\frac{R}{16}\ \ ...(ii)\)
Now, By dividing Equation (i) and Equation (ii)
\(\frac{λ'}{λ}=\frac{R}{4}\times\frac{16}{R}⇒λ'=4λ\)
So, the correct option is (C) : 4λ.