Question 415: Atoms & Nuclei

Question

In Paschen series, wavelength of first line is ' $\lambda_1$ ' and for Brackett series, wavelength of first line is ' $\lambda_2$ ' then ratio $\frac{\lambda_1}{\lambda_2}$ is

Accepted Answer

Using Rydberg's formula,
$$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight) $$
For Paschen series, $\mathrm{n}_1=3$ and $\mathrm{n}_2=4$
$$ \frac{1}{\lambda_1}=R\left(\frac{1}{9}-\frac{1}{16}ight)=\frac{7 R}{16 	imes 9} $$
For Brackett series,
$$ \begin{aligned} & \mathrm{n}_1=4 	ext { and } \mathrm{n}_2=5 \ & \frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{16}-\frac{1}{25}ight)=\frac{9 \mathrm{R}}{16 	imes 25} \ & \frac{\lambda_1}{\lambda_2}=\frac{9 \mathrm{R}}{16 	imes 25} 	imes \frac{16 	imes 9}{7 \mathrm{R}}=\frac{81}{175} \end{aligned} $$

Answer

$\frac{7}{400}$

Answer

$\frac{9}{144}$

Answer

$\frac{81}{175}$

Answer

$\frac{108}{509}$

Atoms & Nuclei

In Paschen series, wavelength of first line is ' \lambda_1 ' and for Brackett series, wavelength of first line is ' \lambda_2 ' then ratio \frac{\lambda_1}{\lambda_2} is

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