Question

Difficulty: EasyRadioactive Decay Law and Half-life

A radioactive sample has an initial activity of 80 Bq80\text{ Bq} and a half-life of 4 days4\text{ days}. What is the activity of the sample that has decayed after 12 days12\text{ days}?

  1. 70 Bq70\text{ Bq}Answer
  2. B
    10 Bq10\text{ Bq}
  3. C
    40 Bq40\text{ Bq}
  4. D
    20 Bq20\text{ Bq}

Answer

The activity that has decayed after 12 days is 70 Bq70\text{ Bq}.
After 12 days, which equals 3 half-lives (12/4=312 / 4 = 3), the remaining activity of the sample is 80×(1/2)3=10 Bq80 \times (1/2)^3 = 10\text{ Bq}. Consequently, the activity that has decayed is the initial activity minus the remaining activity: 80 Bq10 Bq=70 Bq80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}.

Step-by-Step Solution

1
Determine the number of elapsed half-lives
n=tT1/2=12 days4 days=3 half-livesn = \frac{t}{T_{1/2}} = \frac{12\text{ days}}{4\text{ days}} = 3\text{ half-lives}
Dividing the total time elapsed by the half-life period gives the number of decay cycles.
2
Calculate the remaining activity
Aremaining=A0(12)n=80×(12)3=80×18=10 BqA_{\text{remaining}} = A_0 \left(\frac{1}{2}\right)^n = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ Bq}
The remaining quantity decreases by half for each half-life cycle.
3
Calculate the decayed activity
Adecayed=A0Aremaining=80 Bq10 Bq=70 BqA_{\text{decayed}} = A_0 - A_{\text{remaining}} = 80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}
Subtracting the undecayed remaining activity from the initial activity gives the total decayed activity.

Key Concept

Radioactive Decay Law and Half-life
Estimated Time:45s
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