Question

$H$-atoms in ground state $\left(13.6eV\right)$ are excited by monochromatic radiations of photon of energy $12.1eV$. The number of spectral lines emitted in $H$-atom will be:

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

Answer: (C)

$3$

Energy of excited level is $E_n=-13.6+12.1=-1.5eV$
$n^2=\frac{E_1}{E_n}=\frac{13.6}{1.5}=9$
$n=3$
Means electron excited to 3rd orbit and now, transition occurs from $3^{rd}$ to $1^{st}$ orbit.
Spectral lines emitted $=\frac{n(n-1)}{2}=\frac{3\times 2}{2}=3$