# Suppose an Attractive Nuclear Force Acts Between Two Protons Which May Be Written As F=Ce−Kr/R2. Suppose That K = 1 Fermi−1 ,Find the Value Of C. - Physics

Short Note
Sum

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.

#### Solution

By Coulomb's Law, electric force ,

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

Given , k = 1 fermi-1 = 1015

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

∵ Fn = F

⇒ 2.69 × 1026 × C = 9.216

Comparing both the forces, we get

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

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 7 Electric Field and Potential
Q 15.2 | Page 121