Question

Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t = 0. Find the number of active nuclei at time t.

Answer 1

Explanation:

At t = 0, let the atom numbers be $N_{0}$.

Let the radio-active isotopic number be denoted by N at time t.

Then, $N=N_{0} e^{-\lambda} t$

where,

λ = Constant of disintegration  

Therefore, the number of decayed radioactive isotopes = $N_{0}-N_{0} e^{-\lambda t}$

= $N_{0}-N=N_{0}\left(1-e^{-\lambda t}\right)$ …(1)

Decay’s rate, R is shown as $R=\lambda N_{0}$   …(2)

When the value of N0 from equation (2) is substituted in equation (1), we obtain

$\mathrm{N}=\mathrm{R} \lambda(1-\mathrm{e}-\lambda \mathrm{t})=\mathrm{NO}(1-\mathrm{e}-\lambda \mathrm{t})$