Assertion (A): Forces acting between proton-p | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$(A)$ The nuclear force is charge-independent, meaning the strong nuclear force between any two nucleons (proton-proton, proton-neutron, or neutron-ne

Vedclass · Accepted Answer

If both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Vedclass · Answer

$(A)$ The nuclear force is charge-independent, meaning the strong nuclear force between any two nucleons (proton-proton, proton-neutron, or neutron-neutron) is approximately the same at a given distance.However, in the case of two protons, there is an additional electrostatic force of repulsion acting between them due to their positive charges.Since the nuclear force is attractive and the electrostatic force is repulsive, the net force between two protons is $f_{pp} = f_{	ext{nuclear}} - f_{	ext{electrostatic}}$.For proton-neutron or neutron-neutron pairs, there is no electrostatic repulsion, so the net force is simply the attractive nuclear force $(f_{pn} = f_{nn} = f_{	ext{nuclear}})$.Therefore, $f_{pp} Both the Assertion and the Reason are true, and the Reason correctly explains why the net force between protons is smaller.

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.

Similar Questions

The ratio of the volume of an $Al^{27}$ nucleus to its surface area is (Given,$R_0 = 1.2 \times 10^{-15} \ m$):

An $\alpha$-particle consists of:

The radius of ${ }_{29} Cu^{64}$ nucleus in Fermi is (given $R_{0}=1.2 \times 10^{-15} \ m$)

Express unified atomic mass unit $(amu)$ in $kg$.

Find the ratio of the radii of the nuclei $_{13}Al^{27}$ and $_{52}Te^{125}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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.