Question

If $20 \,\mathrm{gm}$ of a radioactive substance due to radioactive decay reduces to $10 \mathrm{gm}$ in 4 minutes, then in what time $80\, \mathrm{gm}$ of the same substance will reduce to $10\,\mathrm{gm}$.
A. $\ln 8$ minutes
B. $\quad \ln 12$ minutes
C. $\quad \ln 16$ minutes
D. $\quad \ln 20$ minutes

Answer 1

Hint:Half-life is the time taken by a radioactive element to disintegrate to its half the value. To solve this problem we have to use the radioactive decay equation which establishes the relationship between the number of atoms/nuclei decay with time t.

Formula used:
Here we use the formula, $N_{(t)}=N_{0}\left(\dfrac{1}{2}\right)^{t / T_{1 / 2}}$.
Concentration of radioactive substance left after $t$ time is given by $N_{(t)}$.
Amount of substance present before starting the decay is represented by $N_{0}$.
Half-life of radioactive substances is represented as $T_{1 / 2}$.

Complete step by step solution:
Radioactive decay is the sudden emission of radiation from the atomic nucleus of a radioactive substance. Radioactive decay is shown by an unstable nucleus. The radioactive decay formula is $N_{(t)}=$ $N_{0} e^{-\lambda t}$. Where $\lambda$ is decay constant. It is also known as disintegration constant. It is the reciprocal value of decay time and its unit is $s^{-1}$.

Half-life is estimated as the radioactive elements disintegrate into half its concentration. Half-life formula is $t_{1 / 2}=\frac{0.693}{\lambda}$. It is clearly given in question that $20 \mathrm{gm}$ substance reduces to its half value in $4 \mathrm{~min}$. Then,
$N_{(t)}=$ $N_{0}\left(\dfrac{1}{2}\right)^{t} / T_{1 / 2} \\ $
Here, $N_{(t)}=10, N_{0}=20$ and $t=4$
$10=20\left(\dfrac{1}{2}\right)^{4 / T_{1 / 2}}$
Solving this we get $T_{1 / 2}=4 \min$
Hence, we find out half-life of this radioactive material. Now we need to find out the time required for a radioactive substance to decay from $80 \mathrm{gm}$ to $10 \mathrm{gm}$.

Substituting the given values to the equation (1), $\quad N_{(t)}=10, N_{0}=80$ and we find out that $T_{1 / 2}=4 \min$. Therefore,
$10=80 \times\left(\dfrac{1}{2}\right)^{\dfrac{t}{4}}$
$\Rightarrow \dfrac{1}{8}=\left(\dfrac{1}{2}\right)^{\dfrac{t}{4}}\\
\Rightarrow \dfrac{1}{2^{3}}=\left(\dfrac{1}{2^{2}}\right)^{\dfrac{t}{4}}\\
\Rightarrow \dfrac{1}{2^{12}}=\dfrac{1}{2^{t}}$
$\therefore t=12 \mathrm{~min}$
Time required for reducing a radioactive material from $80 \mathrm{gm}$ to $10 \mathrm{gm}$ is 12 minutes.

Therefore, the answer is option B.

Notes: Radioactivity is the spontaneous breakdown of unstable atomic nuclei to generate more energetically stable atomic nuclei. Radioactive decay is a highly exoergic, statistically unpredictable, first-order process in which a little quantity of mass is transformed to energy. Because it is a first-order process, each radioactive species has its own half-life, which is the amount of time it takes for a large number of such nuclei to decay to only half their original number.