Question

In a mixture of $H-H{e}^{+}$ gas, H atom and $H{e}^{+}$ ions are excited to their respective first excited states. Subsequently, H atoms transfer its excitation energy to $H{e}^{+}$ ions by collision.

If each hydrogen atom in the ground state is excited by absorbing photons of energy 8.4 eV, 12.09 eV, of energy, then assuming the Bohr model of an atom is applicable the number of spectral lines emitted is equal to:

• A. 5
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is C 3

Energy of electron in $n^{th}$ shell:

$E_n=\frac{–13.12Z^2}{n^2}eV$

$n=1:-13.6eV$

$n=2:-3.4eV$

$n=3:-1.51eV$

$n=4:-0.85eV$

$E(n=3)-E(n=1)=12.09eV$

So if $H$ atoms are excited by $8.4eV$ and $12.09eV$ then the electrons will reach to $n=3$ shell. Then the no. of emitted spectral lines will be equal to $3$.

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