Question 678: Atoms & Nuclei

Question

A freshly prepared sample of a radioisotope of half-life 1386 s has activity 10
3
disintegrations per second. Given that ln2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that will decay in the first 80 s after preparation of the sample is __________.

Accepted Answer

Decay constant, $$\lambda  = {{\ln 2} \over {{T_{1/2}}}} = {{0.693} \over {1386\,s}} = 5 	imes {10^{ - 4}}\,s$$
According to radioactive decay, $$N = {N_0}{e^{ - \lambda t}}$$
$${N \over {{N_0}}} = {e^{ - 5 	imes {{10}^{ - 4}} 	imes 80}}$$ or $${N \over {{N_0}}} = {e^{ - 0.04}}$$
Fraction of nuclei decayed $$ = {{{N_0} - N} \over {{N_0}}} = 1 - {N \over {{N_0}}}$$
$$ = 1 - {e^{ - 0.04}} = 1 - 0.96 = 0.04 = 4\% $$

Atoms & Nuclei

Integer / Numeric Answer Type