Suppose an attractive nuclear force acts between two protons which may be written as F=Ce^{−kr}/r^{2}. 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.

#### Solution

By Coulomb's Law, electric force ,

`F = (Ce ^(-kr))/r^2`

Given , k = 1 fermi^{-1 }= 10^{15} m

Taking

\[r = 5 \times {10}^{- 15} \text{ m }\] , we get

The electrostatic repulsion between the protons,

` therefore F_e = (Kq^2)/r^2`

\[ ⇒ F_e = \frac{9 \times {10}^9 \times \left( 1 . 6 \times {10}^{- 19} \right)^2}{\left( 5 \times {10}^{- 15} \right)^2}\]

= 9.216 N

The strong nuclear force is ,

And nuclear force,` F = Ce^{−kr}/r^{2}

Taking r = 5 × 10^{-15 }m and k = 1 fermi^{−1}, we get

` F_n = (Ce^(-kr))/r^2 = (C xx e^(-10^15 xx 5 xx 10^(-15)))/(5 xx 10^(-15))^2`

\[F_n = \frac{C \times {10}^{- 5}}{\left( 5 \times {10}^{- 15} \right)^2}\]

= 2.69 × 10^{26 } × C

∵ F_{n} = F_{e }

⇒ 2.69 × 10^{26} × C = 9.216

Comparing both the forces, we get

\[C = 3 . 42 \times {10}^{- 26} N \text{m}^2\]