Question

Use the one-dimensional particle-in-a-box model with impenetrable walls and the equation R = R_0*A^(1/3) to estimate the minimum kinetic energy of a nucleon in a nucleus. Express your answer in MeV and in terms of a number 'n' the mass number 'A', and an exponent p, which is the ratio of two integers, resulting in K = n/A^p.

Answer 1

Energy of a particle in a box with impenetrable walls is given by:

Where L = length of the box,

q = a positive integer,

h= 6.626 Js is planck's constant

m = mass of the particle.

In this case, L = 2*R where R is radius of the nucleus

From equation (1) we get:

R₀ = 1.2*10^-15 m

m = 1 u = 1.66*10^-27 kg

Putting these values

Expression (2) gives the kinetic energy of a nucleon inside the nucleus using particle in a one dimensional box model.

For minimum value of kinetic energy, put q = 1, therefore we get, minimum kinetic energy to be