Question
In a radioactive material the activity at time t_1, is A_1 and at a later time t_2, it is A_2. If the decay constant of the material is , then
BITSAT
Physics
Nuclear Physics
2022
BITSAT 2022
BITSAT Physics Question (2022) — Solution
Options
- A. A_1=A_2 e^ - (t_1-t_2 )
- B. A_1=A_2 e^ (t_1-t_2 )
- C. A_1=A_2 (t_2 / t_1 )
- D. A_1=A_2
Answer
A. A_1=A_2 e^ - (t_1-t_2 )
Step-by-step solution
From radioactive decay law, - d N d t N or - d N d t = N Thus, A=- d N d t or A= N A= N_0 e^ - t Where, A_0= N_0 is the activity of the radioactive material at time, t=0 At time, t_1 A_1=A_0 e^ - t_1 At time, t_2 A_2=A_0 e^ - t_2 On dividing Eq. (ii) by Eq. (iii), we have aligned A_1 A_2 & = e^ - t_1 e^ - t_2 =e^ - (t_1-t_2 ) \\ A_1 & =A_2 e^ - (t_1-t_2 ) aligned
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