Q196 / 200 · MHT CET - Medical 2026 (23 April Shift 2) · Physics
At time $t = 0$, the activity of a radioactive substance is $1600$ Bq. At $t= 8$ s, the activity remains $100$ Bq. The activity at $t = 2$ s, the is
- A$400$ Bq
- Correct answerB$800$ Bq
- C$1600$ Bq
- D$3200$ Bq
Solution
The activity of a radioactive substance at time $t$ is given by $A = A_0 \left(\dfrac{1}{2}\right)^{\dfrac{t}{T_{1/2}}}$.
Given $A_0 = 1600$ Bq at $t = 0$.
At $t = 8$ s, $A = 100$ Bq.
$100 = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\left(\dfrac{1}{2}\right)^4 = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\dfrac{8}{T_{1/2}} = 4 \Rightarrow T_{1/2} = 2$ s.
The activity at $t = 2$ s is:
$A = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{2}{2}} = 1600 \times \dfrac{1}{2} = 800$ Bq.
Answer: $800$ Bq
Given $A_0 = 1600$ Bq at $t = 0$.
At $t = 8$ s, $A = 100$ Bq.
$100 = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\dfrac{1}{16} = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\left(\dfrac{1}{2}\right)^4 = \left(\dfrac{1}{2}\right)^{\dfrac{8}{T_{1/2}}}$
$\dfrac{8}{T_{1/2}} = 4 \Rightarrow T_{1/2} = 2$ s.
The activity at $t = 2$ s is:
$A = 1600 \left(\dfrac{1}{2}\right)^{\dfrac{2}{2}} = 1600 \times \dfrac{1}{2} = 800$ Bq.
Answer: $800$ Bq
Hence, the correct answer is option: 'B'.