Q163 / 180 · COMEDK 2021 · Physics
Ba-122 has half-life of $2 \mathrm{~min}$. Experiment has to be done using $B a-122$ and it takes 10 min to set up the experiment. It initially $80 \mathrm{~g}$ at $\mathrm{Ba}-122$ was taken, how much Ba was left when experiment was started?
- Correct answerA$2.5 \mathrm{~g}$
- B$5 \mathrm{~g}$
- C$10 \mathrm{~g}$
- D$20 \mathrm{~g}$
Solution
The half-life of Ba-122 is given as $T_{1/2} = 2 \text{ min}$.
The time elapsed before the experiment starts is $t = 10 \text{ min}$.
The number of half-lives $n$ that have passed is calculated as $n = \dfrac{t}{T_{1/2}} = \dfrac{10}{2} = 5$.
The amount of substance remaining $N$ after $n$ half-lives is given by the formula $N = N_0 \left( \dfrac{1}{2} \right)^n$, where $N_0$ is the initial amount.
Given $N_0 = 80 \text{ g}$, we have $N = 80 \times \left( \dfrac{1}{2} \right)^5$.
$N = 80 \times \dfrac{1}{32} = \dfrac{80}{32} = 2.5 \text{ g}$.
Answer: $2.5 \mathrm{~g}$
The time elapsed before the experiment starts is $t = 10 \text{ min}$.
The number of half-lives $n$ that have passed is calculated as $n = \dfrac{t}{T_{1/2}} = \dfrac{10}{2} = 5$.
The amount of substance remaining $N$ after $n$ half-lives is given by the formula $N = N_0 \left( \dfrac{1}{2} \right)^n$, where $N_0$ is the initial amount.
Given $N_0 = 80 \text{ g}$, we have $N = 80 \times \left( \dfrac{1}{2} \right)^5$.
$N = 80 \times \dfrac{1}{32} = \dfrac{80}{32} = 2.5 \text{ g}$.
Answer: $2.5 \mathrm{~g}$
Hence, the correct answer is option: 'A'.