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

Answer:

$E_1=1.31\times 10^{-12}\ J$

Explanation:

Given that,

Width of a one dimensional potential box, $x=5\times 10^{-15}\ m$

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

$E_n=\dfrac{n^2h^2}{8mx^2}$

h = Planck's constant

m = the mass of the proton

For minimum kinetic energy, n = 1

$E_1=\dfrac{(6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times (5\times 10^{-15})^2}$

$E_1=1.31\times 10^{-12}\ J$

So, the minimum kinetic energy of the neutron is $1.31\times 10^{-12}\ J$. Hence, this is the required solution.