Nuclei and Radioactivity — Practice Quiz

Iron (A ≈ 56)

Step 1: The B/A vs A graph shows that binding energy per nucleon increases from lighter nuclei, reaches a maximum, and then gradually decreases for heavier nuclei. Step 2: The maximum B/A value of approximately 8.8 MeV occurs around A = 56, which corresponds to Iron (Fe). Step 3: Helium has B/A ≈ 7.1 MeV; Uranium has B/A ≈ 7.6 MeV; Hydrogen has the lowest B/A. Step 4: Iron is therefore the most stable nucleus — energy must be supplied to break it apart or to fuse it with anything. Key principle: Iron is the most tightly bound nucleus in nature.

²³⁴₉₀Th

Step 1: In alpha decay, an alpha particle (⁴₂He) is emitted from the parent nucleus. Step 2: Conservation of mass number: 238 − 4 = 234 → daughter has A = 234. Step 3: Conservation of atomic number: 92 − 2 = 90 → daughter has Z = 90, which is Thorium (Th). Step 4: So the daughter nucleus is ²³⁴₉₀Th. Step 5: Option A (Pa, Z=91) would result from beta decay. Option C has wrong A. Option D keeps same Z which is wrong for alpha decay. Key principle: In alpha decay, A decreases by 4 and Z decreases by 2.

10 g

Step 1: Number of half-lives elapsed = Total time / Half-life = 30 / 10 = 3 half-lives. Step 2: After each half-life, the amount becomes half of the previous amount. Step 3: After 1st half-life (10 yr): 80/2 = 40 g. Step 4: After 2nd half-life (20 yr): 40/2 = 20 g. Step 5: After 3rd half-life (30 yr): 20/2 = 10 g. Formula: N = N₀ × (1/2)ⁿ where n = number of half-lives = 80 × (1/2)³ = 80/8 = 10 g.

T₁/₂ = 0.693 / λ

Step 1: From the radioactive decay law N(t) = N₀ e^(−λt), at t = T₁/₂, N = N₀/2. Step 2: Substituting: N₀/2 = N₀ e^(−λT₁/₂). Step 3: Dividing both sides by N₀: 1/2 = e^(−λT₁/₂). Step 4: Taking natural log: ln(1/2) = −λT₁/₂ → −0.693 = −λT₁/₂. Step 5: Therefore T₁/₂ = 0.693/λ. Option D (T₁/₂ = 1/λ) is the mean life (τ), not half-life. Mean life τ = 1/λ = 1.44 × T₁/₂.