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To what series does the spectral line of...

To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the difference between the wave numbers of the following two lines of the Balmer series: `486.1`and `410.2nm`? What is the wavelength of that time?

Text Solution

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Here, `lambda_1=4102Å=4102xx10^(-10)m`
`lambda_2=4861Å=4861xx10^(-10)m`
As Balmer series arises due to jump of electron form higher levels to second orbit, therefore, form
`1/lambda=R(1/(n_1^2)-1/(n_2^2))`, we get
`1/(4102xx10^(-10))=1.097xx10^7[1/(2^2)-1/(n'_1^2)]`
On solving it, we get `n'_1=6`
Again, `1/(4861xx10^(-10))=1.097xx10^7[1/(2^2)-1/(n'_2^2)]`
On solving it, we get `n'_2=4`
for wave number equal to difference in wave number of above two lines, the transition must occur form `n'_1=6` to `n'_2=4`.
`:. 1/(lambda')=R(1/(4^2)-1/(6^2))`
`1/(lambda')=1.097xx10^7(1/16-1/36)`
`=1.097xx10^7xx20/(16xx36)`
`lambda'=(16xx36xx10^-7)/(1.097xx20)m=2.625xx10^-6m`
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