Question

Let $A$ be the mass number, $Z$ be the atomic number and $N$ be the neutron number of a nucleus. Then the statement which is always true, is

• A. $A^2 \ge NZ$
• B. $A \ge 2N$
• C. $A \ge 2Z$
• D. $AN \ge Z^2$

Hint:

Using simple counterexamples like Hydrogen-1 ($A=1, Z=1, N=0$) and Helium-3 ($A=3, Z=2, N=1$) lets you quickly eliminate incorrect options in nuclear physics relations.

Answer 1

The Correct Option is A

Step 1: Note the basic nuclear bookkeeping.
A nucleus has $Z$ protons and $N$ neutrons, and the mass number is simply their total, so $A = Z + N$. All three are zero or positive whole numbers.

Step 2: Decide what we must prove.
We want a relation that can never be broken for any nucleus, including odd cases like very light or very heavy ones.

Step 3: Test the candidate $A^2 \ge NZ$.
Substitute $A = N + Z$, so $A^2 = (N + Z)^2 = N^2 + 2NZ + Z^2$.

Step 4: Compare the two sides.
We need $N^2 + 2NZ + Z^2 \ge NZ$, which rearranges to $N^2 + NZ + Z^2 \ge 0$.

Step 5: Confirm it is always true.
Since $N$ and $Z$ are never negative, every term $N^2$, $NZ$ and $Z^2$ is non-negative, so their sum is certainly at least zero. This holds for every nucleus, no exceptions.

Step 6: Why the others can fail.
Relations like $A \ge 2Z$ or $A \ge 2N$ break for nuclei where protons or neutrons dominate, for example hydrogen-1 has no neutrons. So they are not always true. \[ \boxed{A^2 \ge NZ} \]