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Zorluk: OrtaRadioactivity, Nuclear Reactions, and Half-Life

A sample of iodine-131 (131I^{131}\text{I}), a radioactive isotope used in medical diagnosis, has a half-life of 8 days8\text{ days}. If only 2.5 g2.5\text{ g} of the sample remains active after an elapsed time of 24 days24\text{ days}, what was the initial mass of the sample?

  1. A
    7.5 g7.5\text{ g}
  2. B
    10 g10\text{ g}
  3. 20 g20\text{ g}Cevap
  4. D
    40 g40\text{ g}

Cevap

The initial mass of the iodine-131 sample was 20 g20\text{ g}.
Over an elapsed time of 24 days24\text{ days} with a half-life of 8 days8\text{ days}, exactly 33 half-lives pass (24÷8=324 \div 8 = 3). Since the remaining mass is 2.5 g2.5\text{ g}, working backward requires doubling the mass three times: 2.5 g5.0 g10.0 g20.0 g2.5\text{ g} \rightarrow 5.0\text{ g} \rightarrow 10.0\text{ g} \rightarrow 20.0\text{ g}, giving an initial mass of 20 g20\text{ g}.

Adım Adım Çözüm

1
Determine the number of half-lives (nn) that have elapsed.
n=Total TimeHalf-life=24 days8 days=3 half-livesn = \frac{\text{Total Time}}{\text{Half-life}} = \frac{24\text{ days}}{8\text{ days}} = 3\text{ half-lives}
Dividing total elapsed time by the half-life period gives the total count of half-life cycles.
2
Apply the radioactive decay relationship to calculate initial mass (N0N_0).
N=N0(12)n    2.5 g=N0(12)3=N08N = N_0 \left(\frac{1}{2}\right)^n \implies 2.5\text{ g} = N_0 \left(\frac{1}{2}\right)^3 = \frac{N_0}{8}
Radioactive decay follows an exponential model where remaining amount is reduced by half each cycle.
3
Solve for initial mass (N0N_0).
N0=2.5 g×8=20 gN_0 = 2.5\text{ g} \times 8 = 20\text{ g}
Multiplying the remaining mass by 232^3 reverses the exponential decay process.

Anahtar Kavram

Radioactive Half-Life and Exponential Decay Calculations
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