Question

The half-life for decay of ${}^{14}C$ by $\beta -$emission is 5730 year. The fraction of ${}^{14}C$ decays, in a sample that is 22920 years old, would be-
(A) ${\displaystyle \frac{1}{8}}$
(B) ${\displaystyle \frac{1}{16}}$
(C) ${\displaystyle \frac{7}{8}}$
(D) ${\displaystyle \frac{15}{16}}$

Answer 1

Hint: We will first apply the formula $N=N_0({\displaystyle \frac{1}{2}}{)}^{{\displaystyle \frac{t}{{\displaystyle \frac{t_1}{2}}}}}$, and then we will find out the value of N. Then we will simply find the required answer as per asked by the question. Refer to the solution below.

Complete answer:
> Formula used: $N=N_0({\displaystyle \frac{1}{2}}{)}^{{\displaystyle \frac{t}{{\displaystyle \frac{t_1}{2}}}}}$
> Half-life ($t_{{\displaystyle \frac{1}{2}}}$ symbol) is the period of time required to reduce the initial value of any quantity. In nuclear physics, the term is common for describing how fast unstable atoms are subjected to radioactive decay or how long stable atoms remain. The concept is used to describe exponential or non-exponential decay in a more general way.
 Medical sciences, for instance, relate to the biological half-life of drugs and other chemicals. The half-life opposite is twice as much.
Since we already know the formula for exponential decay is-
$\Rightarrow N=N_0({\displaystyle \frac{1}{2}}{)}^{{\displaystyle \frac{t}{{\displaystyle \frac{t_1}{2}}}}}$
Where, N is considered to be the amount left so the diffraction was created
And $N_0$ is considered to be the initial amount
Now, solving it further by substituting the values given in the question and using the above formula, we get-
$$\begin{gathered} \Rightarrow N=N_0({\displaystyle \frac{1}{2}}{)}^{{\displaystyle \frac{t}{{\displaystyle \frac{t_1}{2}}}}} \\ \Rightarrow N=N_0{\left({\displaystyle \frac{1}{2}}\right)}^{{\displaystyle \frac{22920}{5730}}} \\ \Rightarrow N=N_0{\left({\displaystyle \frac{1}{2}}\right)}^4 \\ \Rightarrow N={\displaystyle \frac{N_0}{16}} \end{gathered}$$
Now, let’s find out the fraction of ${}^{14}C$ decays as per given in the question, will be-
$$\begin{gathered} \Rightarrow N_0-N \\ \Rightarrow N_0-{\displaystyle \frac{N_0}{16}} \\ \Rightarrow {\displaystyle \frac{15}{16}}{N}_0 \end{gathered}$$
Hence, it is clear that option D is the correct option.

Note: Half-life, in radioactivity, the time interval required to decay for half the radioactive sample atomic nuclei or the time interval required to decrease by a half for the number of disintegrations of radioactive material in one second, by spontaneously shifting to another nuclear type by emitting particles and energy.