Figure shows the energy levels P,Q,R,S and G of an atom where G is the ground state. A red line in the emission spectrum of the atom can be obtained by an energy level change from Q to S. A blue line can be obtained by following energy level change

 

$r_n=a_0{n}_2$

$\begin{array}{rr}& {\mathrm{r}}_1={\mathrm{a}}_0=5.3\times {10}^{-11}\mathrm{m}\\ & {\mathrm{r}}_3={\mathrm{a}}_0\left(3\right)2\\ & =5.3\times {10}^{-11}\times 9\\ & =4.77\text{?}\end{array}$

Density of nucleus is constant.

r = 0.5Å = 0.5×10? ¹? m
v = 2.2×10? m/s
I =?
t = 2πr/v
I = e/t = ev/2πr = (1.6×10? ¹? × 2.2×10? )/ (2 × 22/7 × 0.5×10? ¹? )
I ≈ 1.12×10? ³ A = 1.12 mA = 112 × 10? ² mA

$\Delta E=hv\to \left(\frac{1}{n_2^2}-\frac{1}{n_1^2}\right)$

(1) n₁ = 3, n₂ = 2

    $\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{9-4}{36}=\frac{5}{36}$         

(2) n₁ = 4, n₂ = 3

$\left(\frac{1}{3^2}-\frac{1}{4^2}\right)=\frac{16-9}{144}=\frac{7}{144}$        

(3) n₁ = 2, n₂ = 1

$\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{4-1}{4}=\frac{3}{4}$  

(4) n₁ = 5, n₂ = 4

$\left(\frac{1}{4^2}-\frac{1}{5^2}\right)=\frac{25-16}{400}=\frac{9}{400}$