A and B are two substances undergoing radioac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The radioactive decay follows first-order kinetics: $[N]_t = [N]_0 e^{-kt}$,where $k = \frac{\ln 2}{t_{1/2}}$.For substance $A$: $[A]_t = [A]_0 e^

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) The radioactive decay follows first-order kinetics: $[N]_t = [N]_0 e^{-kt}$,where $k = \frac{\ln 2}{t_{1/2}}$.For substance $A$: $[A]_t = [A]_0 e^{-k_A t}$,where $k_A = \frac{\ln 2}{15}$.For substance $B$: $[B]_t = [B]_0 e^{-k_B t}$,where $k_B = \frac{\ln 2}{5}$ and $[B]_0 = 4[A]_0$.We want to find time $t$ such that $[A]_t = [B]_t$.$[A]_0 e^{-(\frac{\ln 2}{15})t} = 4[A]_0 e^{-(\frac{\ln 2}{5})t}$.Dividing both sides by $[A]_0 e^{-(\frac{\ln 2}{5})t}$:$e^{(\frac{\ln 2}{5} - \frac{\ln 2}{15})t} = 4$.Taking the natural logarithm on both sides:$t(\frac{\ln 2}{5} - \frac{\ln 2}{15}) = \ln 4$.$t(\ln 2)(\frac{1}{5} - \frac{1}{15}) = 2 \ln 2$.$t(\frac{3-1}{15}) = 2$.$t(\frac{2}{15}) = 2$.$t = 15 \ min$.

Similar Questions

In Young's double slit experiment,if the maximum intensity is $I$,then the angular position where the intensity becomes $\frac{I}{4}$ is:

Which organ does not participate in fetal circulation?

During the depression in freezing point experiment,an equilibrium is established between the molecules of

If $A$ is any set,then:

When zeolite (hydrated sodium aluminium silicate) is treated with hard water,the sodium ions are exchanged with:

Vedclass Test Series

Exam Paper Generator

Online Exam Module