Answer:
[tex]E_1=1.31\times 10^{-12}\ J[/tex]
Explanation:
Given that,
Width of a one dimensional potential box, [tex]x=5\times 10^{-15}\ m[/tex]
The energy of a particle in one dimensional box is given by :
[tex]E_n=\dfrac{n^2h^2}{8mx^2}[/tex]
h = Planck's constant
m = the mass of the proton
For minimum kinetic energy, n = 1
[tex]E_1=\dfrac{(6.63\times 10^{-34})^2}{8\times 1.67\times 10^{-27}\times (5\times 10^{-15})^2}[/tex]
[tex]E_1=1.31\times 10^{-12}\ J[/tex]
So, the minimum kinetic energy of the neutron is [tex]1.31\times 10^{-12}\ J[/tex]. Hence, this is the required solution.
