How to Calculate Force of Atoms Using the Yukawa-Type Potential Energy Function?

[mopar969]

The Yukawa-type potential energy function for the interaction of a particular pair of neutrons in an atomic nucleus is U(r) = -(a/r)U(sub o) (e^(-r/a)) where U(sub o) = 5 x 10^(-12) J and a = 1.5 x10^(-15) m. Find the force as a function of r, then evaluate the force for r=a and r=3a.

To get force I took Force = -du/dr and got -a(r^-2)(U(subo))(e^(-r/a)). Then I plugged a in for r in the derivative and got -3333.33(e^(-1)). Then I plugged 3a in for r in the derivative and got -0.37037(e^(3)). Is my approach to this problem correct and are my answers correct?

 

[Delphi51]

The approach is right but you must have a mistake in that derivative. It is a product of a/r and e^(-r/a) so you must use the product rule to differentiate and you'll end up with two terms (f'g + fg') in the answer.

 

[mopar969]

Thanks for Identifying that error, please check my new work and answers:
The derivative is ((2a)/(r^2))(Usub o)(e^(-r/a))+(-a/r)(Usub o)((-2(r^2))/a)(e^(-r/a))). When I plugged in a for r I got 2452.53. When I plugged in 3a for r I got 36.8793.

 

[Delphi51]

We aren't quite in agreement on that derivative. In the interest of clarity, forget the constant -a*Uo in front. The rest is
U = r^-1*e^(-r/a)
U = f*g where f = r^-1, g = e^(-r/a)
f ' = -r^-2 g' = -1/a*e^(-r/a)
dU/dr = f 'g + fg'
= -r^-2*e^(-r/a) + r^-1*(-1)/a*e^(-r/a)
=
In particular, your second term has r^1 while mine has r^-1.

 

[mopar969]

When I plugged in a for r I got 2452.52 N. When I plugged in 3a for r I got 73.7587. Are these new numbers correct?

 

[Delphi51]

Yes, I get the same two numbers.

 

[mopar969]

Thank you a lot for all the help.

 

[Delphi51]

Most welcome.