An initial sample contains 4.0×10^6 nuclei of a radioactive isotope. After 10 days only 1.0×10^6 nuclei remain. Assuming exponential decay, calculate the half‑life of the isotope (in days).

An initial sample contains $4.0 \times 1 0^{6}$ nuclei of a radioactive isotope. After $10 days$ only $1.0 \times 1 0^{6}$ nuclei remain. Assuming exponential decay, calculate the half‑life of the isotope (in days).

medium 2 marks Class 12 · Physics · Nuclei

Answer

The half‑life

T_{1/2} = 5, t e x t d a y s

The correct answer is The half‑life $T_{1/2} = 5, t e x t d a y s$ — Thus the half‑life is $5$ days.

Write the decay law: $N = N_{0} (\frac{1}{2})^{t / T_{1/2}}$ .
Substitute $N / N_{0} = 1.0 \times 1 0^{6} /4.0 \times 1 0^{6} = \frac{1}{4}$ and $t = 10 days$ .
Set $l e f t (f r a c 12 r i g h t)^{t / T_{1/2}} = f r a c 14 = (f r a c 12)^{2}$ .
Equate exponents: $t / T_{1/2} = 2$ so $T_{1/2} = t /2 = 10/2 = 5 days$ .
Thus the half‑life is $5$ days.

Common mistake: Treating decay as linear instead of exponential

Concept tested: Radioactive decay law