Physics - Modern Physics Question with Solution | TestHub

Visible light lies in the range, ${\lambda }_1=4000\mathring{A}$ to ${\lambda }_2=7000\mathring{A}$. Energy of photons corresponding to these wavelengths (in $\mathrm{eV}$ ) would be; $$\begin{array}{rl}& E_1=\frac{12375}{4000}=3.09\mathrm{eV}\\ & E_2=\frac{12375}{7000}=1.77\mathrm{eV}\end{array}$$ 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. $$\lambda =\frac{12375}{2.64}\mathring{A}=4687.5\mathring{A}=4.68\times 10^{-7}\mathrm{m}$$ Therefore, (c) is the most correct option.