Question 526: Atoms & Nuclei

Question

In Balmer series, wavelength of first line is ' $\lambda_1$ ' and in Brackett series wavelength of first line is ' $\lambda_2$ ' then $\frac{\lambda_1}{\lambda_2}$ is

Accepted Answer

The wavelength of a line in Balmer series is given by
$$\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}ight)\quad [	ext{for } n=3,4,5, \ldots]$$
where, $R=$ Rydberg constant.
For first line, $n=3$
$$\Rightarrow \quad \frac{1}{\lambda_1}=R\left(\frac{1}{4}-\frac{1}{9}ight) \Rightarrow \lambda_1=\frac{36}{5 R}\quad	ext{.... (i)}$$
The wavelength of a line in Brackett series is given by
$$\frac{1}{\lambda}=R\left(\frac{1}{4^2}-\frac{1}{n^2}ight)\quad 	ext{for } n=5,6,7 \ldots$$
For first line, $n=5$
$$\Rightarrow \quad \frac{1}{\lambda_2}=R\left(\frac{1}{16}-\frac{1}{25}ight) \Rightarrow \lambda_2=\frac{400}{9 R}\quad	ext{.... (ii)}$$
Dividing Eq. (i) by Eq. (ii), we get
$$\frac{\lambda_1}{\lambda_2}=\frac{36}{5 R} 	imes \frac{9 R}{400}=\frac{81}{500}=0.162$$

Answer

0.162

Answer

0.124

Answer

0.138

Answer

0.188

Atoms & Nuclei

In Balmer series, wavelength of first line is ' \lambda_1 ' and in Brackett series wavelength of first line is ' \lambda_2 ' then \frac{\lambda_1}{\lambda_2} is

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