Question

What are the frequency and wavelength of a photon during a transition from n= 5 state to n = 2 state in the He⁺ ion?

Answer 1

Energy between two states = E = (13.6 eV) [$\frac{1}{{{\mathrm{n}}_{\mathrm{f}}}^2}-\frac{1}{{{\mathrm{n}}_{\mathrm{i}}}^2}$]
Energy of n^th state = ​E_n = (-13.6 eV)/n²

So, E₅ = (-13.6eV)/5² = -0.544 eV
E₂ = (-13.6eV)/2² ​= -3.4 eV

Energy during transition from n= 5 state to n = 2 state,
E = (13.6 eV) [$\frac{1}{{{\mathrm{n}}_2}^2}-\frac{1}{{{\mathrm{n}}_5}^2}$]
= E₂-E₅
= -3.4 eV + 0.544 eV​
= -2.856 eV
​Therefore on going from level 5 to level 2, electron loses 2.856 eV of energy.

Converting this to joules gives E = 2.856 $\times$ 1.60 x 10^-19 J/eV = 4.5696 x 10^-19 J

Now, E = $\frac{\mathrm{hc}}{\mathrm{\lambda }}$

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{E}}$ = $\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{4.5696\times {10}^{-19}}$
​ = 4.3527$\times$10^-7 m

The frequency,
$$\begin{gathered} \nu =\frac{c}{\lambda } \\ =\frac{3\times {10}^8}{4.3527\times {10}^{-7}} \\ =6.89s^{-1} \end{gathered}$$