First slide

Radioactivity - Nuclei

Question

The activity of a radioactive sample is 1.6 curie and its half-life is 5 days. Its activity after 20 days will be

Easy

A

0.4 curie
B

0.1 curie
C

0.16 curie
D

0.8 curie

Solution

$N o o f h a l f l i v e s = n = \frac{t}{t_{\frac{1}{2}}} = \frac{20}{5} = 4 A c t i v i t y, A = \frac{A_{0}}{2^{n}} = \frac{1.6}{16} = 0.1 c u r i e$

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