Suppose the magnitude of Nuclear force between two protons varies with the distance between them as shown in figure. Estimate the ratio "Nuclear force/Coulomb force" for

(a) *x* = 8 fm

(b) *x* = 4 fm

(c) *x* = 2 fm

(d) *x* = 1 fm (1 fm = 10 ^{−15}m).

#### Solution

First let us calculate the coulomb force between 2 protons for distance = 8 fm\[F = \frac{K q^2}{r^2}\]

\[ = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(8 \times {10}^{- 15} )^2}\]

\[ = 3 . 6 N\]

\[F_N = 0 . 05 N\]

\[\frac{F_N}{F_C} = \frac{0 . 05}{3 . 6} = 0 . 0138 N\]

For x= 4 fm

\[F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(4 \times {10}^{- 15} )^2}\]

\[ = \frac{23 . 04 \times {10}^{- 29}}{(4 \times {10}^{- 15} )^2}\]

\[ = 14 . 4 N\]

\[ F_N = 1N\]

\[\frac{F_N}{F_C} = \frac{1}{14 . 4} = 0 . 0694 N\]

\[\text{ For }\ x = 2 \text{ fm } \]

\[ F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(2 \times {10}^{- 15} )^2}\]

\[ = 57 . 6 N\]

\[ F_N = 10 N\]

\[\frac{F_N}{F_C} = \frac{10}{57 . 6} = 0 . 173\]

\[\text{ For }\ x = 1 \text{ fm } \]

\[ F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(1 \times {10}^{- 15} )^2}\]

\[ = 230 . 4 N\]

\[ F_N = 1000 N\]

\[\frac{F_N}{F_C} = \frac{1000}{230 . 4} = 4 . 34\]