Question: 38: At any instant, two elements X_{1} and X_{2} have same number of radioactive atoms. If the decay constant of X_{1} and X_{2} are 10\lambda and \lambda respectively. then the time when the ratio of their atoms becomes \frac{1}{e} respectively will be :
(1) \frac{1}{11\lambda}
(2) \frac{1}{9\lambda}
(3) \frac{1}{6\lambda}
(4) \frac{1}{5\lambda}
Answer: Option (2)
Explanation:
Let the number of radioactive atoms of each element at the given instant be N_{0}.
According to the law of radioactive decay, the number of atoms remaining after time
t is given by N=N_{0}e^{-\lambda t}.
For element X_{1} with decay constant 10\lambda,
the number of atoms after time t is N_{1}=N_{0}e^{-10\lambda t}.
For element X_{2} with decay constant \lambda,
the number of atoms after time t is N_{2}=N_{0}e^{-\lambda t}.
The ratio of the number of atoms at time t is given as
\frac{N_{1}}{N_{2}}=\frac{1}{e}.
Substituting the expressions, we get
\frac{N_{0}e^{-10\lambda t}}{N_{0}e^{-\lambda t}}=e^{-9\lambda t}.
Thus, e^{-9\lambda t}=\frac{1}{e}.
Taking logarithm, we obtain -9\lambda t=-1.
Hence, the required time is t=\frac{1}{9\lambda}.
Therefore, the correct option is (2).