Question

Suppose an attractive nuclear force acts between two protons which may be written as F=Ce^−kr/r². 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

By Coulomb's Law, electric force , 

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

Given , k = 1 fermi^-1 = 10¹⁵ 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/r²
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 × 10²⁶  × C  

∵ F_n = F_e 

⇒ 2.69 × 10²⁶ × C = 9.216

Comparing both forces, we get

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