Question

The mean lives of a radioactive substance are $1620$ years and $405$ years for alpha emission and beta emission respectively. The time (in year) during which three fourths of a sample will decay if it is decaying both by alpha emission and beta emission simultaneously is (divide answer by two hundred and write the nearest integer).

Answer 1

Mean life, in radioactivity, the average lifetime of all the nuclei of a particular unstable atomic species. This time span might be considered as the sum of the lifetimes of all the individual unstable nuclei in a sample, divided by the total number of unstable nuclei present. The mean life of specific types of an unstable nucleus is consistently $1.443$ times longer than its half-life (time span needed for a large portion of the unstable nuclei to decay).

Complete step by step solution:
Let at some instant of time $t$ , the number of atoms of the radioactive substance are $N$ . It may decay either by alpha-emission or beta-emission. So, we can write,
${( - \dfrac{{dN}}{{dt}})_{net}} = {( - \dfrac{{dN}}{{dt}})_\alpha } + {( - \dfrac{{dN}}{{dt}})_\beta }$
If the effective decay constant is $\lambda$ , then
$\lambda N = {\lambda _\alpha }N + {\lambda _\beta }N$
$\lambda = {\lambda _\alpha } + {\lambda _\beta } = \dfrac{1}{{1620}} + \dfrac{1}{{405}} = \dfrac{1}{{324}}yea{r^{ - 1}}$
$\dfrac{{{N_0}}}{4} = {N_0}{e^{\lambda t}}$
$- \lambda t = \ln (\dfrac{1}{4}) = - 1.386$
$(\dfrac{1}{{324}})t = 1.386$
$t = 449years$ .

Note:
Alpha decay: Alpha decay or alpha-decay is a type of radioactive decay in which the atomic nucleus emits an alpha particle thereby transforming or decaying into a new atomic nucleus.
Beta decay: Beta-decay occurs in one of the two ways: a) when the nucleus emits an electron and an antineutrino in a process that changes a neutron to a proton b) when the nucleus emits a positron and a neutrino in a process that changes a proton to a neutron.