A sample has initial activity A0. After 12 years, with half-life 3 years, what is the remaining activity?

A sample has initial activity A0. After 12 years, with half-life 3 years, what is the remaining activity?

• A. A0/2
• B. A0/8
• C. A0/32
• D. A0/16

Answer: (D)

Radioactive decay lowers the activity by half every half-life. The remaining activity after time t is A = A0 × (1/2)^(t / T_half). Here the half-life is 3 years and the elapsed time is 12 years, so you have four half-lives (12 ÷ 3 = 4). That gives A = A0 × (1/2)^4 = A0/16. Four halvings in a row produce A0 → A0/2 → A0/4 → A0/8 → A0/16, so the remaining activity is A0/16. Choices corresponding to fewer or more half-lives (like A0/8 or A0/32) don’t fit the 12-year time.

Radioactive decay lowers the activity by half every half-life. The remaining activity after time t is A = A0 × (1/2)^(t / T_half). Here the half-life is 3 years and the elapsed time is 12 years, so you have four half-lives (12 ÷ 3 = 4). That gives A = A0 × (1/2)^4 = A0/16. Four halvings in a row produce A0 → A0/2 → A0/4 → A0/8 → A0/16, so the remaining activity is A0/16. Choices corresponding to fewer or more half-lives (like A0/8 or A0/32) don’t fit the 12-year time.