B)
A hydrogen atom initially in the ground state absorbs a photon
which excites it to n = 4 level. When it gets de-excited, find the
maximum number of lines which are emitted by the atom. Identify
the series to which these lines belong. Which of them has the
shortest wavelength ?
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Answer:
total we will have 6 spectral lines
1 Paschen Series lines is from 4 to 3
2 Balmer series lines from 4 to 2 and 3 to 2
3 Lyman series lines from 4 to 1 , 3 to 1 and 2 to 1
Shortest wavelength out of all above is lyman series line from 4 to 1
Explanation:
As we know that electron reaches to n = 4 due to excitation process
So we will have spectral lines due to de-excitation of electron to lower energy state
so total number of spectral lines are
[tex]^nC_2 = \frac{n(n-1)}{2}[/tex]
here we know that
n = 4
so we will have
[tex]N = \frac{4\times 3}{2} = 6[/tex]
so total we will have 6 lines
1 Paschen Series lines is from 4 to 3
2 Balmer series lines from 4 to 2 and 3 to 2
3 Lyman series lines from 4 to 1 , 3 to 1 and 2 to 1
Shortest wavelength out of all above is lyman series line from 4 to 1
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Topic : Spectral lines in hydrogen spectrum
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