Question

A neutron is confined in a one-demensional potential box of width 5.0 x 10^-15 m. Calculate the minimum kinetic energy of the neutron. If it is confined in a box what is the minimum kinetic energy?

Answer 1

`E_1=1.31* 10^(-12)\ J`

Step-by-step explanation:

Given that,

Width of a one dimensional potential box,
`x=5* 10^(-15)\ m`

The energy of a particle in one dimensional box is given by :

`E_n=(n^2h^2)/(8mx^2)`

h = Planck's constant

m = the mass of the proton

For minimum kinetic energy, n = 1

`E_1=((6.63* 10^(-34))^2)/(8* 1.67* 10^(-27)* (5* 10^(-15))^2)`

`E_1=1.31* 10^(-12)\ J`

So, the minimum kinetic energy of the neutron is
`1.31* 10^(-12)\ J`. Hence, this is the required solution.