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  • The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is:  Option: 1 <img alt="\frac{25}{9}" src="http

The ratio of the longest and shortest wavelengths in Brackett series of hydrogen spectra is:
 

Option: 1

\frac{25}{9}


Option: 2

\frac{17}{5}


Option: 3

\frac{9}{25}


Option: 4

\frac{4}{9}


Answers (1)

best_answer


For Brackett series, \frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{4^2}-\frac{1}{\mathrm{n}^2}\right]

Where \mathrm{n}=5,6,7,8, \ldots \ldots.

For longest wavelength, \mathrm{n}=5

\therefore \frac{1}{\lambda_{\text {Longest }}}=R\left[\frac{1}{4^2}-\frac{1}{5^2}\right]=R\left[\frac{1}{16}-\frac{1}{25}\right]=\frac{9}{400} R

For shortest wavelength, \mathrm{n}=\infty

\therefore \frac{1}{\lambda_{\text {Shortest }}}=R\left[\frac{1}{4^2}-\frac{1}{\infty^2}\right]=\frac{R}{16}

Dividing (ii) by (i), we get


\frac{\lambda_{\text {Longest }}}{\lambda_{\text {Shortest }}}=\frac{\mathrm{R}}{16} \times \frac{400}{9 \mathrm{R}}=\frac{25}{9}

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