Question

The certain radioactive element A, has a half-life = t sec, and in \[\dfrac{t}{2}\] sec, the fraction of the initial quantity of the element so far decayed is nearly
A. \[25\% \]
B. \[29\% \]
C. \[21\% \]
D. \[17\% \]

Answer 1

Hint: The time in which half of the original number of nuclei decay is called as the half-life that is “t sec” in this case. Half of the remaining nuclei decay in the following half-life. Therefore, the number of radioactive nuclei decreases from N to \[\dfrac{N}{2}\] in one half-life and \[\dfrac{N}{4}\] in the next and to $\dfrac{N}{8}$ in the next and so on.

Complete step by step answer:
Radioactive decay shows disappearance of a constant fraction of activity per unit time.
Given in the question is,
Half-life = t sec
Half-life is the time required to decay a sample to \[50\% \] of its initial activity,
So, it means that if X gram is taken initially, then $\dfrac{X}{2}$ gram remains in t seconds.
 \[t = \dfrac{1}{2} = {e^{kt}}\]
Thus, in \[\dfrac{t}{2}\] seconds, fraction decayed will be \[\dfrac{X}{4}\] gram
 \[N\left( t \right) = {N_0}{e^{kt}}\]
 \[A = {e^{kt}}\]
 \[\; \Rightarrow 0.5 = {e^{kt}}\]
 \[K = \dfrac{{ln\left( {0.5} \right)}}{t}\]
So, when \[t = \dfrac{t}{2}\]
 \[A = {e^{\left[ {\dfrac{{In(0.5)}}{t}} \right] \times \left[ {\dfrac{t}{2}} \right]}}\]
 \[ \Rightarrow A = {e^{\left[ {\dfrac{{In(0.5)}}{2}} \right]}}\]
 \[ \Rightarrow A = {e^{0.3465735}}\]
 \[A = 0.707\]
A = amount remaining
So, amount consumed is equal to \[1 - 0.707 = 29.3\% \approx 29\% \]

Therefore, the correct answer is option (B).

Note: Radioactive decay is a spontaneous process. It cannot be predicted exactly for any single nucleus and can only be described statistically and probabilistically i.e. can only give averages and probabilities. Although the decay of individual nuclei happens randomly, it turns out that large numbers of nuclei can be modelled by a mathematical function that predicts the number of radioactive nuclei remaining at a given time is \[N\left( t \right) = {N_0}{e^{kt}}\].