Why D? Only option D uses correct units (U in MeV, r in fm) and matches this behavior.
Others fail because:
The correct answer is (D) : .
In nuclear physics, the potential energy of a pair of nucleons (such as protons or neutrons) typically follows a shape where:
- At very large separations, the potential energy is nearly zero.
- As the nucleons move closer, the potential energy decreases (becomes negative) due to the attractive nuclear force.
- After reaching a minimum, the potential energy increases rapidly at very short distances due to repulsive forces (this is often due to the Pauli exclusion principle for fermions or the electrostatic repulsion between protons).
Graph (D) correctly represents this behavior, showing a sharp attractive region at intermediate distances and a repulsive region at very short distances, which is characteristic of the nuclear force.
The correct answer is (D) : .
Match the LIST-I with LIST-II
LIST-I (Type of decay in Radioactivity) | LIST-II (Reason for stability) | ||
---|---|---|---|
A. | Alpha decay | III. | Nucleus is mostly heavier than Pb (Z=82) |
B. | Beta negative decay | IV. | Nucleus has too many neutrons relative to the number of protons |
C. | Gamma decay | I. | Nucleus has excess energy in an excited state |
D. | Positron Emission | II. | Nucleus has too many protons relative to the number of neutrons |
Choose the correct answer from the options given below:
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below: