Test Index

Atoms and Nuclei Part 3

Section: Physics

Question : 38 of 98

Marks: +1, -0

4^{\text{th}}

Solution: 👈: Video Solution

Number of spectral lines possible when the electron de-excites from the

n^{\text{th}}

energy level to the ground state is

N=\frac{(n_2-n_1)(n_2-n_1+1)}{2}

Here the electron is in the

4^{\text{th}}

energy level, so

n_2=4

and the lowest level is

n_1=1

N=\frac{(4-1)(4-1+1)}{2}=\frac{3\times4}{2}=6

These lines correspond to the transitions

4\to3

4\to2

4\to1

3\to2

3\to1

and

2\to1

Hence the number of spectral lines emitted is

6

So, the correct option is (a).

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