Atoms

According to de-Broglie's hypothesis

$\lambda =\frac{h}{mv}\Rightarrow p=mv=\frac{h}{\lambda }\Rightarrow K=\frac{p^2}{2m}=\frac{h^2}{2m{\lambda }^2}$

Cut-off wavelength of emitted X-ray = $\frac{hc}{K}=hc*\frac{2m{\lambda }^2}{h^2}=\frac{2mc{\lambda }^2}{h}$

Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $\left(\lambda \right)=\frac{1242eV-nm}{6eV}=207nm$

$Frequency=\frac{c}{\lambda }=\frac{3*10^8}{207*10^{-9}}=1.45*10^{9}MHz$

No. of different wavelengths

$=\frac{n*\left(n-1\right)}{2}=\frac{6*5}{2}=15$

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

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

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

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

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

$=2.46*10^{15}Hz$

  $v\propto \frac{z}{n}$ $\frac{}{}$

$\frac{}{}$  $v=k\frac{z}{n}$ [K constant]

for 3rd orbit of He+

${}_{}$ $v_{H{e}^{+}}=\frac{k*2}{3}=\frac{2k}{3}$  - (1)

$$ $v_H=\frac{k*1}{3}=\frac{k}{3}$  - (2)

$\frac{v_{H{e}^{+}}}{v_H}=2:1$

We know

Energy on any orbit is given by,

$E_n=\frac{-13.6}{n^2}{z}^2$

${}_{}\frac{}{}$  $\Rightarrow E_1=\frac{-13.6}{12}*9$ [n1 = 1] (1)

For 3rd orbit

$$ $E_3=\frac{-13.6}{9}*$   [n2 = 3]

E3 = 13.6ev

$\Delta E=E_3-E_1$

= 13.6 – (13.6 * 9)

$\Delta E=8*13.6ev=photonenergy$

$8*13.6ev=\frac{hc}{\lambda }$

$\Rightarrow 8*13.6ev=\frac{1242ev}{\lambda }nm$

$\lambda =11.415*10^9$

= 114.15 * 1010m

Given absorb energy = 10.2 eV

$E_f=\frac{-13.6}{n^2}=-3.4$

$n^2=4$

n = 2

mvr = $\frac{nh}{2\pi }$ $\frac{}{}$

So momentum (mv) = $\frac{nh}{2\pi r}$

According to Rutherford, e- revolves around in nucleus in circular orbit. Thus e- is always accelerating (centripetal acceleration). An accelerating change emits EM radiation and thus e- should loose energy and finally should collapse in the nucleus.

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

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

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

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