Question

A radioactive nucleus is being produced at a constant rate $a$ per second. Its decay constant is $\lambda$. If ${N_0}$ are the number of nuclei at time $t = 0$ , then the maximum number of nuclei possible are?
A.$\dfrac{\alpha }{\lambda }$
B. ${N_0} + \dfrac{\alpha }{\lambda }$
C. ${N_0}$
D. $\dfrac{\lambda }{\alpha } + {N_0}$

Answer 1

In this question given that a radioactive nucleus is being produced at a constant rate. We have to find out when the maximum number of nuclei possible are and the number of nuclei is maximum when rate of decay is equal to rate of information. With the help of this we can find the required answer.

Complete step by step answer:
Question has given the value of $\dfrac{{DN}}{{Dt}}$
It is the rate of decay, and the source is the number of disintegrations per second and it is measured in units which are called Becquerels.
Which means,
$1Bq = 1$disintegration per seconds.
$\dfrac{{DN}}{{Dt}} = a$
And $\dfrac{{DN}}{{Dt}}$is always equal to $N \cdot \lambda$
$\dfrac{{DN}}{{Dt}} = N \cdot \lambda$
Since, we know that,
Maximum number of nuclei will be present when rate of decay $=$rate of production of new nuclei i.e at steady state. So, in this case the maximum number of nuclei will present.
Which is $a = N \cdot \lambda$
Now, we have to calculate value of N, So
$N = \dfrac{a}{\lambda }$
So, the final value is $N = \dfrac{a}{\lambda }$,this is the correct answer.
According to this option A is the correct answer.

Note: The things we have to remember for solving this question is that dn and dt is the rate of disintegration which is measured in per seconds and also remember the units. The material which is containing unstable nuclei is known as radioactive. Knowing all these things we can find the maximum number of nuclei possible. So, these are some important points we have to remember so that we can get the correct answer.