Decay Constant

Problems from IIT JEE

Problem (IIT JEE 2000): Two radioactive materials $\mathrm{X_1}$ and $\mathrm{X_2}$ have decay constants $10\lambda$ and $\lambda$ respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of $\mathrm{X_1}$ to that of $\mathrm{X_2}$ will be $1/e$ after a time

  1. $1/(10\lambda)$
  2. $1/(11\lambda)$
  3. $11/(10\lambda)$
  4. $1/(9\lambda)$

Solution: The populations of $\mathrm{X}_1$ and $\mathrm{X}_2$ after time $t$ are given by, \begin{align} \label{eka:eqn:1} &N_1=N_0e^{-10\lambda t},\\ \label{eka:eqn:2} &N_2=N_0e^{-\lambda t}. \end{align} Divide first equation by second and substitute the values to get, \begin{align} \label{eka:eqn:3} \frac{N_1}{N_2}=\frac{1}{e}=\frac{e^{-10\lambda t}}{e^{-\lambda t}}=e^{-9\lambda t}. \end{align} Take logarithm of above equation and solve to get $t={1}/{(9\lambda)}$.