According to Yukawa theory of Nuclear forces, the attractive force between two Nucleons has the potential V(r)=Ke^-ar/r What will be the Nuclear force?
U(r)=Ke−arr,U(r)=\frac{Ke^{-ar}}r,U(r)=rKe−ar,
F=U′=−arKe−ar−Ke−arr2=−Ke−ar(1+ar)r2.F=U'=\frac{-arKe^{-ar}-Ke^{-ar}}{r^2}=\frac{-Ke^{-ar}(1+ar)}{r^2}.F=U′=r2−arKe−ar−Ke−ar=r2−Ke−ar(1+ar).
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments