Question

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 ⁻¹⁵m).

Answer 1

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
\[F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(4 \times {10}^{- 15} )^2}\]
\[ = \frac{23 . 04 \times {10}^{- 29}}{(4 \times {10}^{- 15} )^2}\]
\[ = 14 . 4 N\]
\[ F_N = 1N\]
\[\frac{F_N}{F_C} = \frac{1}{14 . 4} = 0 . 0694 N\]
\[\text{ For }\ x = 2 \text{ fm } \]
\[ F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(2 \times {10}^{- 15} )^2}\]
\[ = 57 . 6 N\]
\[ F_N = 10 N\]
\[\frac{F_N}{F_C} = \frac{10}{57 . 6} = 0 . 173\]
\[\text{ For }\ x = 1 \text{ fm } \]
\[ F_C = \frac{9 \times {10}^9 \times (1 . 6 \times {10}^{- 19} )^2}{(1 \times {10}^{- 15} )^2}\]
\[ = 230 . 4 N\]
\[ F_N = 1000 N\]
\[\frac{F_N}{F_C} = \frac{1000}{230 . 4} = 4 .34\]