Question

Four thousand active nuclei of a radioactive material are present at $\text{t} = 0.$ After $60$ minutes $500$ active nuclei are left in the sample. The decay constant of the sample is

• A. $\frac{ln \left(2\right)}{60}$ per minute
• B. $\frac{ln \left(2\right)}{20}$ per minute
• C. $20 \, ln \left(2\right)$ per minute
• D. $60 \, ln \left(2\right)$ per minute

Answer 1

Answer: (B)

60 minutes = 3 half lives

$\therefore T_{\frac{1}{2}}=20 \, min$

$\therefore \lambda =\frac{ln 2}{t_{1 / 2}}=\frac{ln 2}{20}per \, min$