Question:

The half life of a radioactive nucleus is $50$ days. The time interval (t$_2-t_1$) between the time t$_2$ when $\frac{2}{3}$ ot it has decayed and the time $t_1$, when $\frac{1}{3}$ of it had decayed is

Updated On: Aug 1, 2022
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The Correct Option is B

Solution and Explanation

According to radioactive decay law $N=N_0e^{-\lambda t}$ where N$_0$ = Number of radioactive nuclei at time t= 0 N = Number of radioactive nuclei left undecayed at any time t $\lambda$= decay constant At time $t_2,\frac{2}{3}$ of the sample had decayed $\therefore N=\frac{1}{3}N_0$ $\quad\therefore \frac{1}{3} N_{0}=N_{0}e^{-\lambda t_{2}} \quad\ldots\left(i\right)$ At time $t_1, \frac{1}{3}$ of the sample had decayed, $\therefore N=\frac{2}{3}N_0$ $\quad\therefore \frac{2}{3}N_{0}=N_{0}e^{-\lambda t_{1}} \quad\ldots\left(ii\right)$ Divide (i) by (ii), we get $\frac{1}{2}=\frac{e^{-\lambda t_2}}{e ^{-\lambda t_1}}$ $\frac{1}{2}=e^{-\lambda t_2}(t_2 -t_1)$ $\lambda(t_2-t_1)=In 2$ $ t_{2}-t_{1}=\frac{In \,2}{\lambda}=\frac{In 2}{\left(\frac{In\, 2}{T_{1 /2}}\right)} \left(\because\lambda=\frac{In\, 2}{T _{1/ 2}}\right)$ $=T_{1 2}=50$ days
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Concepts Used:

Order of Reaction

The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.

Characteristics of the reaction order

  • Reaction order represents the number of species whose concentration directly affects the rate of reaction.
  • It can be obtained by adding all the exponents of the concentration terms in the rate expression.
  • The order of reaction does not depend on the stoichiometric coefficients corresponding to each species in the balanced reaction.
  • The reaction order of a chemical reaction is always defined with the help of reactant concentrations and not with product concentrations.
  • Integer or a fraction form the value of the order of reaction will be there and it can be zero.