The nuclear force between two neutrons in a nucleus is described roughly by the Yukawa potential U(r)=−U0r0re−r/r0, where r is the distance between the neutrons and U0 and r0(≈10−15m) are constants

Part A

Determine the force F(r) .
Express your answer in terms of the variables U0 , r0 , r , and appropriate constants.

F(r) = −U_0e^(−r/r_0) * (1/r+r_0/r^2)

Part B

What is the ratio F(3r0)/F(r0) ?
Express your answer using two significant figures.

[tex]F(\ r) = -\Delta U(r)= U_0r_0 \Delta[\frac{exp(- \frac{r}{r_0} )}{r} ]= U_0r_0[-\frac{exp(-\frac{r}{r_0} )}{rr_0} \bar r - \frac{exp( -\frac{r}{r_0} )}{r^2} \bar r]\\\\F(\ r) = -U_0\ exp (-\frac{r}{r_0} )[\frac{1}{r} +\frac{r_0}{r^2} ]\bar r[/tex]

The ratio of F( 3r0) /F( r0)

F( r0) is calculated as follows;

[tex]F(\ r_0) = -U_0\ exp (-\frac{r_0}{r_0} )[\frac{1}{r_0} +\frac{r_0}{r_0^2} ]\bar r\\\\F(\ r_0) = -0.368U_0[\frac{1}{r_0} +\frac{1}{r_0} ]\bar r\\\\F(\ r_0) = -0.368U_0[\frac{2}{r_0} ]\bar r\\\\F(\ r_0) = -0.736U_0[\frac{1}{r_0} ]\bar r[/tex]

F( 3r0) is calculated as follows;

[tex]F(\ 3r_0) = -U_0\ exp (-\frac{3r_0}{r_0} )[\frac{1}{3r_0} +\frac{r_0}{(3r_0)^2} ]\bar r\\\\F(\ 3r_0) = -0.05U_0[\frac{1}{3r_0} +\frac{r_0}{9r_0^2} ]\bar r\\\\F(\ 3r_0) = -0.05U_0[\frac{1}{3r_0} +\frac{1}{9r_0} ]\bar r\\\\F(\ 3r_0) = -0.05U_0[\frac{4}{9r_0} } ]\bar r\\\\F(\ 3r_0) = -0.022U_0[\frac{1}{r_0} } ]\bar r \\\\[/tex]

Ratio of F( 3r0) to F( r0) is calculated as;

F( 3r0)/F( r0) = (-0.022)/(-0.736)

F( 3r0)/F( r0) = 0.029

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