Question

The uncertainty delta P sets a lower bound on the average momentum of a particle in the nucleus. If a particle's average momentum were to fall below that point, then the uncertainty principle would be violated. Since the uncertainty principle is a fundamental law of physics, this cannot happen. Using delta P = 2.1 x 10^-20 kilogram-meters per second as the minimum momentum of a particle in the nucleus, find the minimum kinetic energy (Kmin) of the particle (in MeV). Use m = 1.7 x 10^-27 kilograms as the mass of the particle. Note that since our calculations are so rough, this serves as the mass of a neutron or a proton.

Answer 1

The minimum kinetic energy of the particle,
`K_(min)=0.808 MeV`

Step-by-step explanation:

In this question we have given,

`\delta p=2.1* 10^(-20)kg.m.s^(-1)`

mass, m=
`1.7* 10^(-27)kg`

we have to find the minimum kinetic energy of the particle,
`K_(min)`=?

We know that momentum and velocity are related by following formula,

`p=m\time v`

Therefore

`2.1* 10^(-20)kg.m.s^(-1)=1.7* 10^(-27) kg* v`

`v=(2.1* 10^(-20)kg.m.s^(-1))/(1.7* 10^(-27) kg)`

`v=1.235* 10^7 ms^(-1)`

Now we know that kinetic energy is given as

`K.E=(mv^2)/(2)`

Therefore minimum kinetic energy is given as

`K_(min)=(1.7* 10^(-27)kg* (1.235* 10^7 ms^(-1))^2)/(2)`

`K_(min)=1.295* 10^(-13) joule`

`K_(min)=1.295* 10^(-13)* 6.242* 10^18 eV`

`K_(min)=0.808* 10^6 eV`

`K_(min)=0.808 MeV`

The minimum kinetic energy of the particle
`K_(min)=0.808 MeV`