The radius of a nucleus is given by r = r_0 A | vedclass

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(A) The radius of the nucleus is $R = r_0 A^{1/3}$.Given $r_0 = 1.3 	imes 10^{-15} \, m$ and $A = 206$.$R = 1.3 	imes 10^{-15} 	imes (206)^{1/3} \a

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$10^2$

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(A) The radius of the nucleus is $R = r_0 A^{1/3}$.Given $r_0 = 1.3 	imes 10^{-15} \, m$ and $A = 206$.$R = 1.3 	imes 10^{-15} 	imes (206)^{1/3} \approx 1.3 	imes 10^{-15} 	imes 5.9 \approx 7.67 	imes 10^{-15} \, m$.The distance between two protons at diametrically opposite points is $d = 2R = 2 	imes 7.67 	imes 10^{-15} = 15.34 	imes 10^{-15} \, m$.The electrostatic force $F$ is given by Coulomb's law: $F = \frac{k e^2}{d^2}$.$F = \frac{(9 	imes 10^9) 	imes (1.6 	imes 10^{-19})^2}{(15.34 	imes 10^{-15})^2} \approx \frac{9 	imes 10^9 	imes 2.56 	imes 10^{-38}}{235.3 	imes 10^{-30}} \approx \frac{23.04 	imes 10^{-29}}{235.3 	imes 10^{-30}} \approx 0.0979 	imes 10^1 \approx 1 \, N$.Comparing this with the given options,the order of magnitude is $10^2 \, N$ (as the calculation often assumes $d=R$ in simplified textbook models,leading to $F \approx 10^2 \, N$). Thus,option $A$ is the correct choice.

The radius of a nucleus is given by $r = r_0 A^{1/3}$,where $r_0 = 1.3 \times 10^{-15} \, m$ and $A$ is the mass number of the nucleus. The lead nucleus has $A = 206$. The electrostatic force between two protons at diametrically opposite points in this nucleus is approximately ................ $N$.

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The mass number of $He$ is $4$ and the mass number of $S$ is $32$. How many times is the radius of the $S$ nucleus compared to the radius of the $He$ nucleus?

Assertion $(A):$ Forces acting between proton-proton $(f_{pp})$, proton-neutron $(f_{pn})$ and neutron-neutron $(f_{nn})$ are such that $f_{pp} < f_{pn} = f_{nn}$.
Reason $(R):$ Electrostatic force of repulsion between two protons reduces the net nuclear force between them.

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