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If the wavelength of the first line of the Balmer series of hydrogen is `6561 Å`, the wavelngth of the second line of the series should be
A. `13122 Å`
B. `3280 Å`
C. `4860 Å`
D. `2187 Å`

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Correct Answer - C
For Balmer series , `n_1=2`,`n_2=3` for `1^st` line and `n_2=4` for second line.
`(lambda_1)/(lambda_2)=(((1)/(2^2)-(1)/(4^2)))/(((1)/(2^2)-(1)/(3^2)))=(3//16)/(5//36)=(3)/(16)xx(36)/(5)=(27)/(20)`
`lambda_2=(20)/(27)lambda_1=(20)/(27)xx6561=4860Å`

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