Question

A sample of four hydrogen atoms, all in the $n^{th}$ excited state, de- excite to the ground state, and shows a maximum of six different spectral lines. Hence the value of $n$ is:

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

$n^{th}$ excited state means that the electron is in the $(n+1{)}^{th}$ energy level. Given, the sample of these $4-H$ atoms shows a total of six different spectral lines, while returning to the ground level, that is possible only when
$\frac{(n+1)n}{2}=6\Rightarrow n=3$
The $3^{rd}$ excited state means the $4^{th}$ orbit.
The respective transitions will be $4\to 1,4\to 2,4\to 3,3\to 2,3\to 1,2\to 1$
Hence, option (a) is correct.