Revision: Quantum Mechanics Applied Physics 1 BE Civil Engineering Semester 1 (FE First Year) University of Mumbai

Heisenberg'suncertainty principle states that one can not measure position and
moment of the moving particle exactly.Thus,the inaccuracies Δx and Δp in the
simultaneous determination of the position'x' and momentum 'p'respectively of a particle are related as 

Δx•Δp≥ħ 

Where ħ=h/2π, h being Planck's constant.

Non existence of electron inside the nucleus.

If the electromagnetic is inside the nucleus of radius of the other of10^-15m,the maximum
uncertainty in the position of electron will be of the order of its radius

∴Δxmax=10^-15

From the limiting condition of Heisenberg's uncertainty principle,

Δxmax•Δpmin≥ħ

Δpmin=ħ/Δxmax=6.63×10^-34/2×3.14×10-15=1.055×10^-19kg-m/sec.

Now,Δpmin= mΔvmin

Hence,   Δvmin=Δpmin/m=1.055×10^-19/9.1×10^-31

=1.159×10¹¹m/s>c

As    Δvmin<v     v>1.159×10¹¹ m/a>c

Therefore, the electron behaves as are lativistic particle.

The relativistic energy of the electronis

E=√mo ²c⁴+p²c²

Since,the actual momentum of the electronp >>Δpmin,p2c2>>mo
2c2,therest

mass energy of the electron the value of which is 0.511MeV.Hence,

E=pc

Assuming p=Δpmin,the least energy that an electron should posses within a nucleus

is given by

Emin=Δpmin•c=1.055x10^-19x3x10⁸

=3.165x10^-11J  Emin=3.165×10^-11/1.6×10^-19=197MeV

Inreality, the only source of generation of electron within a nucleus is the process of
decay.The maximum kinetic energy possessed by the electrons during β-decay is about100 KeV.This shows that an electron can not exist within a nucleus

Numerical Solutlon :

Data :v=350m/sec,Δv/v=0.01

Formula: Δx•Δp≥ħ

Calculations: Δx.m.Δv≥ħ

Δv=350×0.01/100=0.035
Δx≥ħ/mΔv≥6.63×10^-34/2×3.14×9.1×10^-31×0.035

≥3.314×10^-3m

Answer: Minimum uncertainty in position is 3.314×10^-3m