Question

Electron cannot exist nucleus using hesinerbg uncertanity principal prove that

Answer 1

According to the Heisenberg;s uncertainity princle, it is impossible to simultaneously determine the postion and velocity of an electron with great accuracy.

Numerically, If we consider Δv to be the uncertainity in velocity ,Δp to be the uncertainity in momentum, Δx to be uncertainity in position and h to be the planck’s constant (6.626∗10−34Js),

The above law can be expressed numerically as

ΔxΔp≥h4π

Where m is the mass of electron,

Substituting, p=mv we get,

ΔxΔv≥h4πm

Now, if we consider the radius of the atomic nucleus to be 10−15m and mass of electron, ′m′ to be 9.1∗10−31 kg, we get

10−15Δv≥6.626∗10−344∗3.14∗9.1∗10−31

Δv≥6.626∗10−344∗3.14∗9.1∗10−31∗10−15

Δv≥5.79∗1010 m/s

Calulating Δv, we get a value of

5.79∗1010 m/s which contradicts the theory of relativity. What this means is that if an electron exists in the nucleus, it has to travel with a speed of 5.79∗1010 m/s. An object can only travel faster than light if it has no mass but electrons do have mass, hence they can’t travel faster than the speed of light which is precisely 299,792,458 m/s.

Hence, an electron can’t exist in the nucleus.