Mass of electron is low as compared to proton. Hence when both enter into the uniform magnetic region, the electron will move in a circular path with higher frequency on the opposite direction to the current.

Daughter nuclei are more stable than parent nuclei. 

Bohr's postulate:

An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds, to a certain energy level.

Electrons revolve in a circular orbit. Centripetal force is provided by electrostatic force between electron and proton.

Given as,

$\frac{{\mathrm{mv}}^2}{\mathrm{r}} = \frac{1}{4{\mathrm{\pi r}}_0}\frac{{\mathrm{e}}^2}{{\mathrm{r}}^2}$

The motion of an electron in a circular orbit is restricted in such a manner that its angular momentum is an integral multiple of h/2π, Thus

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

3.

An electron may jump spontaneously from one orbit (energy level E₁) to the other orbit (energy level E₂) (E₂ > E₁); then the energy change AE in the electron jump is given by Planck’s equation

∆E = E₂-E₁ = hv

Where h = Planck’s constant.

And v = frequency of light emitted.

When the electron jumps from nth higher orbit to pth lower orbit it emits energy in form of the photon.

E_n - E_p = hv

According to de'Broglie hypothesis

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} = \frac{\mathrm{h}}{\mathrm{mv}} \\ \mathrm{n\lambda } = 2\mathrm{\pi r} \\ \mathrm{n}\frac{\mathrm{h}}{\mathrm{mv}} = 2\mathrm{\pi r} \\ \mathrm{nh} = 2\mathrm{\pi mvr} \\ \mathrm{L} = \frac{\mathrm{nh}}{2\mathrm{\pi }} \end{gathered}$$