Question

An electron in hydrogen atom falls from sixth excited state ti first excited state.Total possible lines emmitted in the visible region are

Answer 1

Answer: The total possible number of spectral lines that are emitted in the visible region is 15.

Explanation:

We are given, an electron in H2 atom falls from 6th excited state i.e., N2 = 6 to its first excited state i.e., N1 = 1.

Now, we need to find the no. of possible lines that are emitted during the falling off of electrons from 6th excited state to first excited state.

This can be done by using the Probability formula which is given as,  

$\frac{(N2 - N1) * (N2 - N1 + 1)}{2}$ …… (i)

 

Thus, by substituting the given values in the formula (i), we get

The possible number of spectral lines produced in the visible region as,

= $\frac{(6 - 1) * (6 - 1 + 1)}{2}$  

= $\frac{5 * 6}{2}$

= 5 * 3

= 15

Answer 2

Answer:

see the attached image

Explanation:

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