Atoms

  $\Delta E=13.6\left(\frac{1}{1^2}-\frac{1}{5^2}\right)=13.6*\frac{24}{25}eV$

$\Rightarrow \frac{hc}{\lambda }=13.6*\frac{24}{25}eV..........\left(1\right)$          

With the help of conservation of linear momentum, we can write

$\frac{h}{\lambda }=m_H{v}_H\Rightarrow \frac{hc}{\lambda }=c{m}_H{v}_H\Rightarrow v_H=\frac{\frac{hc}{\lambda }}{c{m}_H}=\frac{13.6*\frac{24}{25}*1.6*10^{-19}}{3*10^8*1.67*10^{-27}}=4.17m/s$           

$h{f}_1=R\left(\frac{1}{1^2}-\frac{1}{3^2}\right)=R*\frac{8}{9}$

$h{f}_2=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=R*\frac{3}{4}$

$\frac{f_1}{f_2}=\frac{8/9}{3/4}=\frac{32}{27}$

f2 = f1 * $\left(\frac{27}{32}\right)$

$=\left(2.92*10^{15}\right)*\left(\frac{27}{32}\right)$

$=2.46*10^{15}Hz$

If n₂ -> n₁ in H (Z = 1) gives $\lambda$ then $z{n}_2\to z{n}_1$  give same $\lambda$  H-like ion for He⁺ ion (z = 2)

(hc/λ) = 13.6Z² (1/1² - 1/4²) = 13.6Z² (3/4) ; [λ = (4/3)k]

β line of Paschen
(hc/λ') = 13.6Z² (1/9 - 1/25)
λ = k * (9*25/16) ; λ' = (3/4)λ * (9*25/16) = (3³*5²/2? )λ

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

First excited state $\Rightarrow n=2$ $$

${\mathrm{T}}_1=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{4}\mathrm{e}\mathrm{V}$

Second excited state $\Rightarrow \mathrm{n}=3$ $$

${\mathrm{T}}_2=13.6\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2}=-\frac{13.6}{9}\mathrm{e}\mathrm{V}$

${\mathrm{T}}_1:{\mathrm{T}}_2=\frac{1}{4}:\frac{1}{9}=9:4$

In the frame of car

$\mathrm{N}=\mathrm{m}\mathrm{g}$

and  $\mathrm{f}=\mathrm{m}\mathrm{a}$

$\mathrm{f}\le \mu \mathrm{N}$

$\Rightarrow \mathrm{a}\le \mu \mathrm{g}$

$\Rightarrow \mathrm{a}\le 1.5{\mathrm{m}\mathrm{s}}^{-2}$

${\mathrm{a}}_{\mathrm{m}\mathrm{a}\mathrm{x}}=1.5{\mathrm{m}\mathrm{s}}^{-2}$

From the given relations:
mv²/r = |dU/dr| = 4|U? |r³ . (1)
mvr = nh / 2π . (2)
By combining equations (1) and (2), we can derive the radius r:
r = ( (nh)² / (4π√* (U? )*) )¹/³ ⇒ r ∝ n²/³

The Balmer series lies in the visible region of the electromagnetic spectrum.