Question

A neutron is confined in a 10-fm-diameter nucleus.

If the nucleus is modeled as a one-dimensional rigid box, what is the probability that a neutron in the ground state is less than 2.0 fm from the edge of the nucleus?
P = %

Answer 1

The value is
`P  =  9.72%`

Step-by-step explanation:

From the question we are told that

The diameter is
`d  =  10 \ f m`

The position of the neutron of consider is between
`0 \to 2 \  fm`

Generally the wave function representing the position of the neutron is mathematically represented as

`\Phi (x)  =  \sqrt{ (2)/(d) }  sin [(\pi x)/(d) ]`

Generally probability that a neutron in the ground state is less than 2.0 fm from the edge of the nucleus (with the assumption that the diameter of both nucleus and neutron are symmetrical )is mathematically represented as

`=2 *   \int\limits^2_0 { |\sqrt{ (2)/(d) }  sin [(\pi x)/(d) ]|^2} \, dx`

`= (4)/(d)  \int\limits^2_0 sin [(\pi x)/(d) ] \, dx`

From Trigonometry identity

`[sin(\theta)]^2  =  (1- cos (2\theta ))/(2)`

So

`= (4)/(d)  \int\limits^2_0 (1 -  cos [(2\pi x)/(d) ])/(2) \, dx`

`=(2)/(d) [x -  (sin [(2\pix)/(y) ])/([(2\pi)/(d) ]) ]\  |\left  2} \atop {0}} \right.`

`= (2)/(d) [2 -  (sin[(2\pi (2))/(d) ])/((2\pi)/(d) ) ]`

`= (2)/(10) [2 -  (sin[(2\pi (2))/(10) ])/((2\pi)/(d) ) ]`

`P =0.0972`

Converting to percentage

`P  =  100 * 0.0972`

`P  =  9.72%`