Question

A hydrogen atom is in ground state. In order to get six lines in its emission spectrum, wavelengths of incident radiation should be,

• A. $800\mathring{\text{A}}$
• B. $825\mathring{\text{A}}$
• C. $975\mathring{\text{A}}$
• D. $1025\mathring{\text{A}}$

Answer 1

The correct option is C $975\mathring{\text{A}}$

The number of spectral lines emitted, during transition is, $\frac{n(n-1)}{2}$
$\Rightarrow 6=\frac{n}{2}(n-1)$
$\Rightarrow 12=n^2-n$
$\Rightarrow n^2-n-12=0$
$\Rightarrow n^2-4n+3n-12=0$
$\Rightarrow n(n-4)+3(n-4)=0$
$\Rightarrow (n+3)(n-4)=0$
$n=-3,4$
Considering positive value, we have $n=4$
Now, $\frac{1}{\lambda }=R(\frac{1}{1^2}-\frac{1}{4^2})$
$\Rightarrow \lambda =\frac{16}{15R}=975\mathring{\text{A}}$

Hence, $\left(C\right)$ is the correct answer.