A sample contains 5.0×10^5 nuclei of a radioactive isotope. After 24 days, only 6.25×10^4 nuclei remain. Which of the following is the half‑life of the isotope?

A sample contains $5.0 \times 1 0^{5}$ nuclei of a radioactive isotope. After $24$ days, only $6.25 \times 1 0^{4}$ nuclei remain. Which of the following is the half‑life of the isotope?

easy MCQ 2 marks Class 12 · Physics · Nuclei

Answer

8 days

The correct answer is 8 days — Answer: 8 days.

Write the exponential decay law: $N = N_{0} e^{- l amb d a t}$ .
Express the decay constant in terms of half‑life: $l amb d a = l n 2/ T_{1/2}$ .
Set up the ratio: $N / N_{0} = e^{- l amb d a t} = e^{- (l n 2/ T_{1/2}) t} .$
Take natural logarithm: $l n (N / N_{0}) = - (l n 2/ T_{1/2}) t$ .
Rearrange to solve for $T_{1/2}$ : $T_{1/2} = - (l n 2), t / l n (N / N_{0})$ .
Compute $N / N_{0} = 6.25 \times 1 0^{4} /5.0 \times 1 0^{5} = 0.125$ .
Calculate $l n (0.125) = - 2.07944$ and $t = 24$ days.
Plug in: $T_{1/2} = - (0.6931 \times 24) / (- 2.07944) = 8.0$ days.
Hence the half‑life is $8$ days.
Answer: 8 days.

Common mistake: Using

T_{1/2} = tl n (N_{0} / N)

instead of correct formula

Concept tested: Half‑life from exponential decay