Paragraph: In a mixture of H-H^ + gas (He^ + . is singly ionised He...

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 $E_1 and E_2$. $\Delta E_{43}=-3.4-(-6.04)=2.64 \mathrm{eV}Wa$velength 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.