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  • If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å , the wavelength of the second line of the series should beOption: 1</strong

If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å , the wavelength of the second line of the series should be

Option: 1

13122 Å


Option: 2

3280 Å


Option: 3

4860 Å


Option: 4

2187 Å


Answers (1)

best_answer

The wavelength of spectral line in Balmer series is
\mathrm{given\: by \:\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right] }

For first line of Balmer series, n = 3
\mathrm{\Rightarrow \frac{1}{\lambda_1}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 R}{36} ; \, \text{For second line\; }n=4}.
\mathrm{\Rightarrow \frac{1}{\lambda_2}=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3 R}{16} }
\mathrm{\therefore \frac{\lambda_2}{\lambda_1}=\frac{20}{27} \Rightarrow \lambda_1=\frac{20}{27} \times 6561=4860 \AA}

Posted by

shivangi.bhatnagar

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