A certain radioactive material can undergo th | vedclass

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

Question

(C) The rate of decay for each process is given by $\frac{dN_1}{dt} = -\lambda_1 N$,$\frac{dN_2}{dt} = -\lambda_2 N$,and $\frac{dN_3}{dt} = -\lambda_3

Vedclass · Accepted Answer

$\lambda_{eff} = \lambda_1 + \lambda_2 + \lambda_3$

Vedclass · Answer

(C) The rate of decay for each process is given by $\frac{dN_1}{dt} = -\lambda_1 N$,$\frac{dN_2}{dt} = -\lambda_2 N$,and $\frac{dN_3}{dt} = -\lambda_3 N$.The total rate of decay is the sum of the individual rates: $\frac{dN}{dt} = \frac{dN_1}{dt} + \frac{dN_2}{dt} + \frac{dN_3}{dt}$.Substituting the expressions,we get: $\frac{dN}{dt} = -\lambda_1 N - \lambda_2 N - \lambda_3 N = -(\lambda_1 + \lambda_2 + \lambda_3) N$.Comparing this with the general decay equation $\frac{dN}{dt} = -\lambda_{eff} N$,we find that the effective decay constant is $\lambda_{eff} = \lambda_1 + \lambda_2 + \lambda_3$.

$A$ certain radioactive material can undergo three different types of decay,each with a different decay constant $\lambda_1$,$\lambda_2$,and $\lambda_3$. Then the effective decay constant is:

Similar Questions

Beta rays emitted by a radioactive material are

Alpha particles are

$A$ radioactive element ${}_{92}^{242}X$ emits two $\alpha$-particles,one electron,and two positrons. The product nucleus is represented by ${}_{P}^{234}Y$. The value of $P$ is $..................$

The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module