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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Physics
Chapter
Nuclear Physics

BITSAT 2022 · Physics · Nuclear Physics

In a radioactive material the activity at time t_1 , is A_1 and at...

In a radioactive material the activity at time t1t_1, is A1A_1 and at a later time t2t_2, it is A2A_2. If the decay constant of the material is λ\lambda, then

Options

  1. AA_1=A_2 e^{-\lambda\left(t_1-t_2\right)}
  2. BA_1=A_2 e^{\lambda\left(t_1-t_2\right)}
  3. CA1=A2(t2/t1)A_1=A_2 \left(t_2 / t_1\right)
  4. DA1=A2A_1=A_2

Correct answer

A. A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}

Step-by-step solution

From radioactive decay law, -dNdt∝\frac{d N}{d t} \propto N  or  −dNdt=λ\text{\text{ or } }-\frac{d N}{d t}=\lambda N Thus, A=-dNdt or  A=λN⇒A=λN0e−λtWhere,\frac{d N}{d t} \text{\text{ or } } A=\lambda N \Rightarrow \quad A=\lambda N_0 e^{-\lambda t} Where, A_0=\lambda N_0istheactivityoftheradioactivematerialattime,is the activity of the radioactive material at time,t=0Attime,At time,t_1 \quad A_1=A_0 e^{-\lambda t_1}Attime,At time,t_2 \quad A_2=A_0 e^{-\lambda t_2}OndividingEq.(ii)byEq.(iii),wehaveOn dividing Eq. (ii) by Eq. (iii), we haveA1A2=e−λt1e−λt2=e−λ(t1−t2)⇒A1=A2e−λ(t1−t2)\begin{aligned} \quad\frac{A_1}{A_2}& =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda\left(t_1-t_2\right)} \\ \Rightarrow \quad \quad A_1 & =A_2 e^{-\lambda\left(t_1-t_2\right)} \end{aligned}
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