A radioactive sample has an initial activity of 80\text{ Bq} and a half-life of 4\text{ da… — JAMB UTME

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

$70\text{ Bq}$ Cevap
B
$10\text{ Bq}$
C
$40\text{ Bq}$
D
$20\text{ Bq}$

Cevap

The activity that has decayed after 12 days is

70\text{ Bq}

After 12 days, which equals 3 half-lives (

12 / 4 = 3

), the remaining activity of the sample is

80 \times (1/2)^3 = 10\text{ Bq}

. Consequently, the activity that has decayed is the initial activity minus the remaining activity:

80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}

Adım Adım Çözüm

Determine the number of elapsed half-lives

n = \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.

Calculate the remaining activity

A_{\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.

Calculate the decayed activity

A_{\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.

Anahtar Kavram

Radioactive Decay Law and Half-life

Tahmini Süre:45s

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