A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

(a) From the relation, 

$\mathrm{v} = \frac{\mathrm{c}}{\mathrm{n}}\mathrm{\alpha }, \mathrm{where} \mathrm{\alpha } = \frac{2\mathrm{\pi } {\mathrm{Ke}}^2}{\mathrm{ch}} = 0.0073$

Therefore, the speed of electron in hydrogen atom in n=1, 2 and 3 is given by, 
                    $$\begin{gathered} {\mathrm{v}}_1 = \frac{3\times {10}^8}{1}\times 0.0073 = 2.19\times {10}^6 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_2 = \frac{3\times {10}^8}{2}\times 0.0073 = 1.095 \times {10}^6 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_3 =\frac{3\times {10}^8}{3}\times 0.0073 =7.3\times {10}^5 \mathrm{m}/\mathrm{s}. \end{gathered}$$

(b) Orbital period in each of the level is given by, 

$$\begin{gathered} \mathrm{T} = \frac{2\mathrm{\pi r}}{\mathrm{v}} \\ \mathrm{As} {\mathrm{r}}_1 = 0.53\times {10}^{-10}\mathrm{m} \end{gathered}$$  

${\mathrm{T}}_1 = \frac{2\mathrm{\pi }\times 0.53\times {10}^{-10}}{2.19\times {10}^6}= 1.52\times {10}^{-16}\mathrm{s}$ ' 

As,  ${\mathrm{r}}_2 = 4 {\mathrm{r}}_1 \mathrm{and} {\mathrm{v}}_2 = \frac{1}{2}{\mathrm{v}}_1$ 

$\therefore$         $$\begin{gathered} {\mathrm{T}}_2 = 8 {\mathrm{T}}_1 \\ = 8 \times 1.52 \times {10}^{-16}\mathrm{s} \\ = 1.216 \times {10}^{-15}\mathrm{s} \end{gathered}$$ 

Also as, ${\mathrm{r}}_3 = 9 {\mathrm{r}}_1 \mathrm{and} {\mathrm{v}}_3 = \frac{1}{3}{\mathrm{v}}_1$ 
$\therefore$ $$\begin{gathered} {\mathrm{T}}_3 = 27 {\mathrm{T}}_1 = 27 \times 1.52 \times {10}^{-16}\mathrm{s} \\ = 4.1 \times {10}^{-15} \mathrm{s}. \end{gathered}$$ 

T₁, T₂ and T₃ are respectively the orbital time period for n₁, n₂ and n₃ .