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  • If the wavelength of the first line of the Balmer series of hydrogen atom is , the wavelength of the

If the wavelength of the first line of the Balmer series of hydrogen atom is 6561\, \mathrm{ \AA}, the wavelength of the second line of the series should be:

Option: 1

13122\, \mathrm{\AA}


Option: 2

3280\, \mathrm{\AA}


Option: 3

4860\, \mathrm{ \AA}


Option: 4

2187\, \mathrm{\AA}


Answers (1)

best_answer

Wavelength of spectral line in Balmer series is given by

\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{\mathrm{n}^2}\right]

For first line of Balmer series, n = 3

\therefore \quad \frac{1}{\lambda_1}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5 \mathrm{R}}{36}

For second line of Balmer series n = 4

\frac{1}{\lambda_2}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 \mathrm{R}}{16}

\therefore \quad \frac{\lambda_2}{\lambda_1}=\frac{20}{27}

\therefore \quad \lambda_1=\frac{20}{27} \times 6561=4860 \AA

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Riya

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