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प्रश्न
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.
उत्तर
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
∵ Fn = Fe
⇒ 2.69 × 1026 × C = 9.216
Comparing both forces, we get
\[C = 3 . 42 \times {10}^{- 26} N \text{m}^2\]
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