Question

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 ?​

Answer 1

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

$^nC_2 = \frac{n(n-1)}{2}$

here we know that

n = 4

so we will have

$N = \frac{4\times 3}{2} = 6$

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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