Paragraph: In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised…
Paragraph:
In a mixture of $\mathrm{H}-\mathrm{H}^{+}$gas $\left(\mathrm{He}^{+}\right.$is singly ionised $\mathrm{He}$ atom), $\mathrm{H}$ atoms and $\mathrm{He}^{+}$ions are excited to their respective first excited states. Subsequently, $\mathrm{H}$ atoms transfer their total excitation energy to $\mathrm{He}^{+}$ions (by collisions). Assume that the Bohr model of atom is exactly valid.Question:
The wavelength of light emitted in the visible region by $\mathrm{He}^{+}$ions after collisions with $\mathrm{H}$ atoms is
$6.5 \times 10^{-7} \mathrm{~m}$
$5.6 \times 10^{-7} \mathrm{~m}$
$4.8 \times 10^{-7} \mathrm{~m}$
$4.0 \times 10^{-7} \mathrm{~m}$
Solution
Visible light lies in the range, $\lambda_1=4000 Å$ to $\lambda_2=7000 Å$.
Energy of photons corresponding to these wavelengths (in $\mathrm{eV}$ ) would be;
$
\begin{aligned}
& E_1=\frac{12375}{4000}=3.09 \mathrm{eV} \\
& E_2=\frac{12375}{7000}=1.77 \mathrm{eV}
\end{aligned}
$
From energy level diagram of $\mathrm{He}^{+}$atom we can see that in transition from $n=4$ to $n=3$, energy of photon released will lie between $E_1$ and $E_2$.
$
\Delta E_{43}=-3.4-(-6.04)=2.64 \mathrm{eV}
$
Wavelength of photon corresponding to this energy.
$
\lambda=\frac{12375}{2.64} Å=4687.5 Å=4.68 \times 10^{-7} \mathrm{~m}
$
Therefore, (c) is the most correct option.
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