A neutron is confined in a 10fm-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 2.0fm from the edge of the nucleus?

Short Answer

Expert verified

The probability is9.72%.

Step by step solution

01

Given information 

We know that ,

Diameter =10fm

Neutron at positon =2fm

We have to find the probability of the ground state.

02

Simplify

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

φ(x)=2LsinπxL

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

P=200.2fmφ(x)2dxP=200.2fm2Lsin2πxLdxP=4L00.2fm(1-cos2πxL)dxP=4Lx-sin2πxLL2π02fmP=410fm2fm-sin2π×2fm10fm10fm2πP=0.0972P=9.72%

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