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The atomic spectrum of hydrogen contains...

The atomic spectrum of hydrogen contains a series of four lines havinng wavelengths 656.5, 486.3, 434.2 and 410.3 nm. Determine the wavelength of the next line in the same series `[R_(H)=109678cm^(-1)]`.

Text Solution

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As the given wavelengths lie in the visible region, they should belong to the Balmer series, For balmer series, `n_(1)=2`. The value of `n_(2)` for the shortest wavelength (410.3nm) can be determined using the equation,
`overline(v)=(1)/(lamda)=R_(H)((1)/(2^(2))-(1)/(n_(2)^(2)))`
or, `(1)/(410.3xx10^(-7)cm)=109678cm^(-1)((1)/(4)-(1)/(n_(2)^(2)))`
or, `(1)/(n_(2)^(2))=(1)/(4)-(1)/(410.3xx10^(-7)xx109678) or, (1)/(n_(2)^(2))=0.25-0.22`
or, `n_(2)^(2)=(1)/(0.03)=33.33" "therefore n_(2)=6`
Thus, the next line is obtained as a result of the transition of an electron from `n_(2)=7 ` to `n_(1)=2`.
`therefore (1)/(lamda)=109678((1)/(2^(2))-(1)/(7^(2)))cm^(-1)=25181cm^(-1)`
or, `lamda=(1)/(25181)=3.971xx10^(-5)cm=397.1nm`
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