The atomic spectrum of hydrogen is found to contain a series of lines of wavelengths 656.46, 486.27, 434.17 and 410.29 nm. The wavelength (in nm) of the next line in the series will be….
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Correct Answer – -397.13
The given wavelengths lie in the visible region. Hence, they are expected to belong to Balmer series. Thus, `n_(1) = 2`. Let us calculate `n_(2)` for the shortest wavelength, viz., 410.29 nm
`(1)/(lamda) = R_(H) ((1)/(2^(2)) – (1)/(n_(2)^(2)))`
`(1)/(410.29 xx 10^(-7) cm) = 109,67 cm^(-1) ((1)/(4) – (1)/(n_(2)^(2)))`
This on solving gives `n_(2) = 6`
Thus, the next line will be obtained for jump from
`n_(2) = 7 ” to ” n_(1) = 2`, so that
`(1)/(lamda) = 109,677 cm^(-1) ((1)/(2^(2)) – (1)/(7^(2)))`
`= 109,677 xx ((1)/(4) – (1)/(49)) cm^(-1)`
`= 109,677 xx (45)/(196) cm^(-1) = 25180.9 cm^(-1)`
or `lamda = (1)/(25180.9 cm^(-1)) = 397.1 xx 10^(-7) cm`
`= 397.13 nm`