Question

Suppose an attractive nuclear force acts between two protons which may be written as F=Ce^−kr/r². (a) Write down the dimensional formulae and appropriate SI units of C and k. (b) Suppose that k = 1 fermi⁻¹ and that the repulsive electric force between the protons is just balanced by the attractive nuclear force when the separation is 5 fermi. Find the value of C.

Answer 1

(a) Given, nuclear force of attraction, $F=C\frac{e^{-Kr}}{r^2}$
Here e^−Kr is just a pure number, i.e. a dimensionless quantity. So,
$$\begin{gathered} \left[C\right]=\left[F\right]\times \left[r^2\right] \\ \left[C\right]=\left[{\mathrm{MLT}}^{-2}\right]\times \left[{\mathrm{L}}^2\right] \end{gathered}$$
$\left[\mathrm{C}\right]=\left[{\mathrm{ML}}^3{\mathrm{T}}^{-2}\right]$
$\mathrm{And}\left[K\right]=\frac{1}{\left[r\right]}=\left[{\mathrm{L}}^{-1}\right]$

(b)By Coulomb's Law, electric force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{e^2}{r^2}$
Taking $r=5\times {10}^{-15}\mathrm{m}$, we get
$F=\frac{9\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{{\left(5\times {10}^{-15}\right)}^2}$
And nuclear force, F = Ce^−kr/r²
Taking r = 5 × 10^-15 m and k = 1 fermi⁻¹, we get
$F=\frac{C\times {10}^{-5}}{{\left(5\times {10}^{-15}\right)}^2}$
Comparing both the forces, we get
$C=3.4\times {10}^{-26}\mathrm{N}-{\mathrm{m}}^2$