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Zorluk: OrtaRadioactive Decay Law and Half-life

A medical sample contains a radioactive isotope with a half-life of 8 days8\text{ days}. If the initial activity of the sample is 160 kBq160\text{ kBq}, what is the activity corresponding to the portion of the sample that has decayed after 32 days32\text{ days}?

  1. 150 kBq150\text{ kBq}Cevap
  2. B
    10 kBq10\text{ kBq}
  3. C
    40 kBq40\text{ kBq}
  4. D
    20 kBq20\text{ kBq}

Cevap

150 kBq150\text{ kBq}
In 32 days32\text{ days}, a sample with an 8-day8\text{-day} half-life undergoes 44 complete half-lives (32/8=432 / 8 = 4). The remaining activity is 160×(1/2)4=10 kBq160 \times (1/2)^4 = 10\text{ kBq}. The activity that has decayed is the initial activity minus the remaining activity: 160 kBq−10 kBq=150 kBq160\text{ kBq} - 10\text{ kBq} = 150\text{ kBq}.

Adım Adım Çözüm

1
Calculate the total number of half-lives (nn) that have elapsed.
n=Total time (t)Half-life (T1/2)=32 days8 days=4n = \frac{\text{Total time } (t)}{\text{Half-life } (T_{1/2})} = \frac{32\text{ days}}{8\text{ days}} = 4
Determining how many half-life periods fit into the total elapsed time.
2
Calculate the remaining activity (AA) of the sample.
A=A0(12)n=160 kBq×(12)4=160×116=10 kBqA = A_0 \left(\frac{1}{2}\right)^n = 160\text{ kBq} \times \left(\frac{1}{2}\right)^4 = 160 \times \frac{1}{16} = 10\text{ kBq}
Applying the radioactive decay law to find the undecayed fraction.
3
Calculate the activity corresponding to the decayed amount.
Adecayed=A0−A=160 kBq−10 kBq=150 kBqA_{\text{decayed}} = A_0 - A = 160\text{ kBq} - 10\text{ kBq} = 150\text{ kBq}
Subtracting the remaining activity from the initial activity gives the decayed activity.

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Radioactive Decay Law and Half-life
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Radioactive Decay Law and Half-life

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