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  4. The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectrum is

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Q. The ratio of the longest to shortest wavelengths in Brackett series of hydrogen spectrum is

Rajasthan PETRajasthan PET 2001

Solution:

$\frac{1}{\lambda_{\max }} =R\left[\frac{1}{4^{2}}-\frac{1}{5^{2}}\right]=\frac{9}{25 \times 16} R \\ \text { and } \frac{1}{\lambda_{\min }} = R\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}\right]=\frac{R}{16} \\ \therefore \quad \frac{\lambda_{\max }}{\lambda_{\min }} =\frac{25}{9} $