A radioactive nuclide is produced at a consta | vedclass

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

Question

(C) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate: $\frac{dN}{dt} = n - \lambda N$.Rearranging t

Vedclass · Accepted Answer

$N = \frac{n}{\lambda} + \left( N_0 - \frac{n}{\lambda} \right) e^{-\lambda t}$

Vedclass · Answer

(C) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate: $\frac{dN}{dt} = n - \lambda N$.Rearranging the terms,we get $\frac{dN}{n - \lambda N} = dt$.Integrating both sides with limits from $t = 0$ $(N = N_0)$ to $t = t$ $(N = N)$:$\int_{N_0}^{N} \frac{dN}{n - \lambda N} = \int_{0}^{t} dt$.Using the substitution method,$-\frac{1}{\lambda} [\ln(n - \lambda N)]_{N_0}^{N} = t$.$-\frac{1}{\lambda} \ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = t$.$\ln\left( \frac{n - \lambda N}{n - \lambda N_0} \right) = -\lambda t$.$\frac{n - \lambda N}{n - \lambda N_0} = e^{-\lambda t}$.$n - \lambda N = (n - \lambda N_0) e^{-\lambda t}$.$\lambda N = n - (n - \lambda N_0) e^{-\lambda t}$.$N = \frac{n}{\lambda} - \left( \frac{n}{\lambda} - N_0 \right) e^{-\lambda t} = \frac{n}{\lambda} + \left( N_0 - \frac{n}{\lambda} \right) e^{-\lambda t}$.

$A$ radioactive nuclide is produced at a constant rate of $n$ per second (e.g.,by bombarding a target with neutrons). If the number of nuclei at $t = 0$ is $N_0$,the number of nuclei $N$ at time $t$ is given by (where $\lambda$ is the decay constant):

Similar Questions

Carbon dating is best suited for determining the age of fossils if their age in years is of the order of

The activity of an element becomes $\frac{1}{64}$ of its original value in $60 \ s$. Then the half-life period is ............ $s$.

$A$ radioactive isotope has a half-life of $T$ years. How long will it take the activity to reduce to $(a)$ $3.125\%$ $(b)$ $1\%$ of its original value?

$A$ radioactive sample with a half-life of $1$ month has the label: "Activity $= 2 \, \mu Ci$ on $1-8-1991$." What will be its activity two months earlier in $\mu Ci$?

$A$ radioactive element has a half-life period of $800$ years. After $6400$ years,what fraction of the initial amount will remain?

Vedclass Test Series

Exam Paper Generator

Online Exam Module