Initially assume that an electron is a part of a nucleus. The size of a nucleus is about 1 fermi = $10^{(-15)}$m if an electron is confined within a nucleus the uncertainty in its position must not be greater than the dimension of the nucleus i.e., $10^{(-15)}$m. hence,

$∆x_m= 10^{(-15)}m$

From the limiting condition of Heisenberg’s uncertainty principle given in the equation it can be written as

$∆x_m.∆p_mi=ħ$

$∆p_{mi}= ħ/(∆x_{mi} )= (6.63 ×10^{(-34)})/(2×3.14×10^{(-15)} ) = 1.055×10^{(-19)}kg-m/sec$

Now , $∆p_mi=m∆v_mi$

Hence ,
$∆v_mi= (∆p_mi)/m= (1.055×10^{(-19)})/(9.1 ×10^{(-31) }) = 1.159 ×10^{11}m/s \lt c$

$As , ∆v_mi \lt v , v \gt 1.159 ×10^{11}m/s \gt c$

Therefore the electron inside the nucleus behaves as a relativistic particle.

The relativistic energy of the electron is $ E = √(m_{o}^2 c^4 )+√p^2 c^{24}$

Since the actual momentum of the electron $p\gt\gt ∆p_mi.p^2.c^2≫m_o^2.c^2$, the rest mass energy of the electron the value of which is 0.511 MeV. Hence,
E= pc

Assuming p = ∆p_mi the least energy that an electron should posses within a nucleus is given by

$E_mi= ∆p_mi.c $

= $1.055 ×10^{(-19)}×3×10^8$

= $3.165 ×10^{(-11)}J$

$ E_mi= (3.165×10^{(-11)})/(1.6×10^{(-19)} ) = 197 MeV.$

In reality the only source of generation of electron with in a nucleus is the process of β-decay. The maximum kinetic energy possessed by the electrons during β-decay is about 100KeV. This shows that an electron can not exist within a nucleus.

NUMERICAL:-

Given Data :- V = 300m/sec , ∆v/v=0.01 %

Formula :- ∆x.∆p ≥ ħ

Calculations :- ∆x.m.∆p ≥ ħ

∆v =300×0.01/100 = 0.03

$ ∆x ≥ħ/m∆v ≥ (6.63 ×10^{(-34)})/(2×3.14×9.1×0.03×10^{(-31)} )$

$≥ 3.8 ×10^{(-3)}$

Therefore uncertainty in position = $ 3.8 ×10^{(-3)}m$