Question

The mean life is $\tau$ for a radioactive nucleus. If the number of decays per unit time is $n$ at $t=0$, then the decays between $0$ and $t$ is:
A. $nT{e}^{-{\displaystyle \frac{t}{T}}}$
B. $n\left(1-e^{-{\displaystyle \frac{t}{T}}}\right)$
C. $nT\left(1-e^{-{\displaystyle \frac{t}{T}}}\right)$
D. $n{e}^{-{\displaystyle \frac{t}{\tau }}}$

Answer 1

Hint: Radioactivity is a nuclear phenomenon. Unstable nuclei which emit radiation, as a result there are two types of unstable nuclei which are found in our common environment, such as uranium and thorium. An atom is stable only when forces among the particles that make nucleus. If an atom is unstable then these forces are unbalanced.

Complete step by step answer:
The mean life in the question is given as $\tau$. We know mean life,
$\tau ={\displaystyle \frac{1}{\lambda }}---\left(1\right)$
where $\lambda$ is the decay constant.
The number of decays per unit time is given $n$ which means the total amount of nuclei that is present is $n$. According to the given question the number of decay times at time $t=0$ is $n$. The number of decay times at time $t=t$ be $N$. From Radioactive Decay Law we find that,
${\displaystyle \frac{dN}{dt}}=-\lambda N$
Arranging the equation we get,
${\displaystyle \frac{dN}{N}}=-\lambda dt$
Integrating the equation we get,
$\int {\displaystyle \frac{dN}{N}}=-\int \lambda dt$
As we have to find the decays between $t=0$ to $t$, according to the question the amount of nuclei at time $t=0$ is $n$ and at $t=t$ be $N$. Thus, we find the limits of the equation,
${\int }_n^N{\displaystyle \frac{dN}{N}}=-{\int }_0^t\lambda dt$

Now we get,
${\left[\log N\right]}_n^N=-\lambda {\left[t\right]}_0^t$
Putting the limits in the equation we get,
$\log N-\log n=-\lambda t$
$\Rightarrow \log {\displaystyle \frac{N}{n}}=-\lambda t$
Simplifying the equation and removing $\log$ we get,
${\displaystyle \frac{N}{n}}=e^{-\lambda t}$
By cross-multiplication we get,
$N=n{e}^{-\lambda t}$
Putting the value of $\lambda$ in terms of $\tau$ from equation $\left(1\right)$ we get,
$\therefore N=n{e}^{-{\displaystyle \frac{t}{\tau }}}$

So, the correct option is D.

Additional information: Mean life in radioactive means, average life time of all nuclei of a particular unstable atomic species. Number of disintegrations per second is equal to the number of atoms. The average life of a radioactive is given by, ratio of total life time to all individual parent atoms.

Note: Nuclear radiation occurs in the other form of emission of protons or neutrons.In the nucleus, atoms are positively charged whereas electrons are negatively charged. Emission of radiation in the form of particles or high energy photons, results in the nuclear reaction.