Question

With Heisenberg’s uncertainty principle prove that electron cannot survive in nucleus. An electron has a speed of 300m/sec. with uncertainty of 0.01% . find the accuracy in its position.

Answer 1

Initially assume that an electron is a part of a nucleus. The size of a nucleus is about 1 fermi = 10^-15m 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^-15m. hence,

∆x_m = 10^-15m 

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 >c"`

As , ∆v_mi < v , v> 1.159 ×  10¹¹m/s >c 

Therefore the electron inside the nucleus behaves as a relativistic particle.
The relativistic energy of the electron is `E = sqrt( m_o^2c^4+√p^2c^2`

Since the actual momentum of the electron p>> ∆p_mi .p².c² ≫ m²_o.c² , 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 ×10-¹⁹×3×10⁸ 
= 3.165 ×10^-11J 
`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 ≥ h/(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^-3m