Question

A radioactive nucleus is being produced at a constant rate $\alpha$ 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. $N_{0}+\frac{\alpha}{\lambda}$
• B. $N_0$
• C. $\frac{\lambda}{\alpha}+N_{0}$
• D. $\frac{\alpha }{\lambda }$

Answer 1

The Correct Option is D

Maximum number of nuclei will be present, when rate of decay = rate of formation $\lambda, N=\alpha, N=\frac{\alpha}{\lambda}$

Answer 2

 Here, dn/dt is quantified in units called Becquerels and the source of the term is the number of disintegrations per second.

dn/dt = α, according to the given problem.

dn/dt is also equal to Nλ.

Therefore, α = Nλ

When the rate of decay equals the rate of generation of new nuclei, or at steady state, the maximum number of nuclei will be present. Therefore, the greatest number of nuclei will be present in this scenario.

So, N = α/λ

The number of nuclei undergoing decay and the rate of decay per unit of time is directly proportional to the total number of nuclei present in the given sample of the radioactive material, according to the law of radioactive decay, whenever the material experiences a decay (alpha, beta, or gamma decay).

Mathematical representation of the Law of radioactive decay:

Where,

ΔN = number of nuclei in the sample that undergoes radioactive decay.

N = the total number of nuclei in the sample

Δt =unit time

Simplifying the equation:  ΔN = λN. Δt 

Where λ is the radioactive decay constant.