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MHT CET · Physics · Laws of Motion

In a radioactive material the activity at time \(t_{1}\) is \(R_{1}\) and at a later time \(t_{2}\), it is \(R_{2}\). If the decay constant of the material is \(\lambda\), then

  1. A \(R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}\)
  2. B \(R_{1}=R_{2} e^{\lambda\left(t_{1}-t_{2}\right)}\)
  3. C \(R_{1}=R_{2}\left(t_{2} / t_{1}\right)\)
  4. D \(R_{1}=R_{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}\)

Step-by-step Solution

Detailed explanation

The decay rate \(R\) of a radioactive material is the number of decays per second. From radioactive decay law.
\(-\frac{d N}{d t} \propto N \quad \text { or }-\frac{d N}{d t}=\lambda N\)
Thus, \(\quad R=-\frac{d N}{d t}\)
Or \(R=\lambda N \quad\) or \(\quad R=\lambda N_{0} e^{-\lambda t} \ldots\)..(i)
where \(R_{0}=\lambda N_{0}\) is the activity of the radioactive material at time \(t=0\). At time \(t_{1}, \quad R_{1}=R_{0} e^{-\lambda t_{1}}\)
At time \(t_{2}, \quad R_{2}=R_{0} e^{-\lambda t_{2}}\)
Dividing Eq. (ii) by (iii), we have
or
\(\begin{array}{l}\frac{R_{1}}{R_{2}}=\frac{e^{-\lambda t_{1}}}{e^{-\lambda t_{2}}}=e^{-\lambda\left(t_{1}-t_{2}\right)} \\
R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}\end{array}\)