Question

When an electron is in 4th excited energy state in hydrogen atom, the possible number of emission spectral lines is

Answer 1

HELLO THERE!

An electron is present in the 4th excited state, this means n = 5 (as n = 1 is the ground state, n = 2 is the 1st excited state, n = 3 is the 2nd excited state, hence n = 5 is the 4th excited state).

There's a formula for number of spectral lines, and it's mentioned below:

$\frac{(n_{2}-n_{1})(n_{2}-n_{1}+1)}{2}$

Here,

$n_{2} = 5\\n_{1} = 0$

So, the formula for number of spectral lines comes out to be:

$\frac{n(n+1)}{2}\\\\= \frac{5\times6}{2}\\\\= 15$

That's the answer!

Thanks!