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
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Hint :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 Answer:
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.
Complete Step By Step Answer:
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.
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