Question

What is the ratio of wave number of first line (lowest energy line) of Balmer series of H atomic spectrum to first line of its Brackett series?

• A. 5:1
• B. 5:0.81
• C. 5:1.75
• D. 5:2.7

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
[4pt] The wavenumber \(\left(\bar{\nu}\right)\) of spectral lines in the hydrogen emission spectrum is given by the Rydberg formula. Different spectral series are formed when the electron falls to different lower energy levels \(n_1\). \[ \text{Balmer series: } n_1=2 \] \[ \text{Brackett series: } n_1=4 \] Step 2: Formula Used
[4pt] The Rydberg equation is: \[ \bar{\nu}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \] For the first line of any series: \[ n_2=n_1+1 \] Thus, \[ \text{Balmer first line: } n_1=2,\; n_2=3 \] \[ \text{Brackett first line: } n_1=4,\; n_2=5 \] Step 3: Wavenumber of first Balmer line
[4pt] \[ \bar{\nu}_{\text{Balmer}} = R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \] \[ = R_H\left(\frac{1}{4}-\frac{1}{9}\right) \] \[ = R_H\left(\frac{9-4}{36}\right) \] \[ = R_H\left(\frac{5}{36}\right) \] Step 4: Wavenumber of first Brackett line
[4pt] \[ \bar{\nu}_{\text{Brackett}} = R_H\left(\frac{1}{4^2}-\frac{1}{5^2}\right) \] \[ = R_H\left(\frac{1}{16}-\frac{1}{25}\right) \] \[ = R_H\left(\frac{25-16}{400}\right) \] \[ = R_H\left(\frac{9}{400}\right) \] Step 5: Calculate the ratio
[4pt] \[ \text{Ratio} = \frac{\bar{\nu}_{\text{Balmer}}}{\bar{\nu}_{\text{Brackett}}} \] \[ = \frac{\frac{5}{36}}{\frac{9}{400}} \] \[ = \frac{5}{36}\times\frac{400}{9} \] \[ = \frac{2000}{324} \] \[ = \frac{500}{81} \] Final Answer: \[ \boxed{\frac{500}{81}} \]