Question

A neutron is confined in a $10\mathrm{fm}$-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 $2.0\mathrm{fm}$ from the edge of the nucleus?

Short answer

The probability is$9.72\%.$

Step-by-step solution

Given information 

We know that ,

Diameter =$10\mathrm{fm}$

Neutron at positon =$2\mathrm{fm}$

We have to find the probability of the ground state.

Simplify

In general for the ground state for any particle the wave function is given by,

$\mathrm{\phi }\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \frac{\mathrm{\pi x}}{\mathrm{L}}$

Then, the probability is defines as below with factor of 2 due to neutron is distributed symmetrical around the nucleus.

$$\begin{gathered} \mathrm{P}=2{\int }_0^{0.2\mathrm{fm}}{\left|\mathrm{\phi }\left(\mathrm{x}\right)\right|}^2\mathrm{dx} \\ \mathrm{P}=2{\int }_0^{0.2\mathrm{fm}}\frac{2}{\mathrm{L}}{\sin }^2\frac{\mathrm{\pi x}}{\mathrm{L}}\mathrm{dx} \\ \mathrm{P}=\frac{4}{\mathrm{L}}{\int }_0^{0.2\mathrm{fm}}(1-\cos \frac{2\mathrm{\pi x}}{\mathrm{L}})\mathrm{dx} \\ \mathrm{P}=\frac{4}{\mathrm{L}}{\left[\mathrm{x}-\sin \left(\frac{2\mathrm{\pi x}}{\mathrm{L}}\right)\frac{\mathrm{L}}{2\mathrm{\pi }}\right]}_0^{2\mathrm{fm}} \\ \mathrm{P}=\frac{4}{10\mathrm{fm}}\left[2\mathrm{fm}-\sin \left(\frac{2\mathrm{\pi }\times 2\mathrm{fm}}{10\mathrm{fm}}\right)\frac{10\mathrm{fm}}{2\mathrm{\pi }}\right] \\ \mathrm{P}=0.0972 \\ \mathrm{P}=9.72\% \end{gathered}$$