Question

For beta-minus decay, which statement is TRUE?

• A. The daughter nuclide atomic mass \((A_D)\) is more than that of the parent nuclide atomic mass \((A_P)\)
• B. The daughter nuclide atomic number \((Z_D)\) is the same as that of the parent nuclide atomic number \((Z_P)\)
• C. The daughter nuclide neutron number \((N_D)\) is less than that of the parent nuclide neutron number \((N_P)\)
• D. The daughter nuclide neutron number \((N_D)\) is the same as that of the parent nuclide neutron number \((N_P)\)

Hint:

In beta-minus decay a neutron becomes a proton: A stays, Z goes up by one, N drops by one.

Answer 1

The Correct Option is C

Method: bookkeeping the three nuclear tallies.

Beta-minus emission is driven by a single neutron converting to a proton. Write down what each conserved or changed quantity does, then test the four claims.

Nucleon (mass) number $A$: a neutron becoming a proton keeps the total nucleon count fixed, so $A_D = A_P$. That kills option (A), which claims the daughter is heavier in mass number.

Proton (atomic) number $Z$: one proton is gained, so $Z_D = Z_P + 1$. That kills option (B), which claims $Z$ is unchanged.

Neutron number $N$: one neutron is lost, so $N_D = N_P - 1$. This immediately makes option (D) false (it claims $N$ is unchanged) and makes option (C) true, since $N_P - 1$ is indeed less than $N_P$.

A quick consistency check: $A = Z + N$ must still hold. Parent has $A_P = Z_P + N_P$; daughter has $(Z_P+1) + (N_P-1) = Z_P + N_P = A_P$, which matches the unchanged mass number. Everything is consistent, and the only correct statement is that the daughter has fewer neutrons than the parent.

\[\boxed{N_D = N_P - 1 < N_P}\]