Question

The counting rate observed from a radioactive source at t second was ${{N}_{0}}$and at 4t second it was $\dfrac{{{N}_{0}}}{16}$, the count rate observed at $(\dfrac{11}{2})t$will be
A-$\dfrac{{{N}_{0}}}{128}$
B-$\dfrac{{{N}_{0}}}{64}$
C-$\dfrac{{{N}_{0}}}{32}$
D- None of the above

Answer 1

This is the problem of radioactive decay. Radioactivity states that a radioactive sample decays with time by the emission of appropriate rays. This is a natural phenomenon and with time the initial number of nuclei starts decreasing.

Complete step by step answer:
It is given that the initial number of undecayed nuclei is ${{N}_{0}}$ at time = t
Using the formula of radioactive decay, $N={{N}_{0}}{{[\dfrac{1}{2}]}^{n}}$
Here N is the number of undecayed nuclei at time = t and ${{N}_{0}}$is the number of nuclei at the starting.
As per the question,
At time = 4t,

& \dfrac{{{N}_{0}}}{16}={{N}_{0}}{{[\dfrac{1}{2}]}^{n}} \\\ &\Rightarrow \dfrac{1}{4}={{[\dfrac{1}{2}]}^{n}} \\\ &\Rightarrow n=4 \\\ \end{aligned}$$ So, 3t times is equivalent to four half-lives. Hence, one half-life is equal to $$\dfrac{3t}{4}$$ The given time is $$\dfrac{11t}{2}-t=\dfrac{9t}{2}$$which is equivalent to 6 half-lives, $$\begin{aligned} & N={{N}_{0}}{{[\dfrac{1}{2}]}^{6}} \\\ &\therefore N =\dfrac{{{N}_{0}}}{64} \\\ \end{aligned}$$ **Hence, the correct option is (B).** **Additional information:** Radium and polonium are examples of radioactive substances. The time in which half of the original number of nuclei decay is defined as the half-life. Half-life is an important parameter which describes the stability of the sample. Some radioactive samples have half-lives of thousands of years. **Note:** Nuclear decay is an example of a purely statistical process. The activity of a radioactive substance is defined as the rate of the number of disintegrations per second. Its SI unit is Becquerel. Other units include Curie and Rutherford. They are bigger units.