Question

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

$(\text{Take:}hc=12425\text{eV}\dot{A})$

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

Answer 1

The correct option is A $975\dot{A}$

The number of possible emission lines is given by,

$N=\frac{n(n-1)}{2}$

So, To get six possible emission lines,

$\Rightarrow 6=\frac{n}{2}(n-1)$

$\Rightarrow 12=n^2-n$

$\Rightarrow n^2-n-12=0$

$\Rightarrow n^2-4n+3n-12=0$

$\Rightarrow n(n-4)+3(n-4)=0$

$\Rightarrow (n+3)(n-4)=0$

$\Rightarrow n=-3,4$

As, $n$ should be positive, $n=4$

So, the electron has to get excited to third excited state to emit $6$ emission lines.

$\Rightarrow$ Energy to be supplied to the hydrogen atom at the ground state to get excited to the $n=4$ is,

$\Delta E=E_4-E_1$

$\Rightarrow \Delta E=-0.85\text{eV}-(-13.6\text{eV})$

$\Rightarrow \frac{hc}{\lambda }=12.75\text{eV}$

$\Rightarrow \lambda =\frac{12425\text{eV}\dot{A}}{12.75\text{eV}}$

$\Rightarrow \lambda =975\dot{A}$

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