Question

Number of spectral lines of Lyman series obtain in He+ ion sample when electron de- excited from 4th excited state is?

Answer 1

Answer:

The answer is 6 spectral lines.

Explanation:

The general formula for the number of spectral lines emitted is

N = (n2 - n1)(n2 - n1 +1)/2

N = (4 -1)(4-1 + 1)/2

N = 3 x 4/2

N = 12/2

N = 6

Thus there are 6 transitions and 6 spectral lines possible. Which are

4 ---> 3

4 ---> 2

4 ---> 1

3 ---> 2

3 ---> 1

2 ---> 1

Answer 2

Number of spectral lines is N = 6

Explanation:

If the electron jumps from $n_2$ = 4 to $n_1$ = 1 then following are the transitions possible.

$4 \rightarrow3\\4\rightarrow2\\4\rightarrow1\\3 \rightarrow2\\3\rightarrow1\\2 \rightarrow1$

Hence there are 6 transitions and hence 6 spectral lines possible.

The general formula for the number of spectral lines emitted is -

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

So,

$N = \frac{(4 - 1)\times (4 - 1 + 1)}{2}$

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

$N = \frac{12}{2}$

$N = 6$