In a radioactive material,the activity at tim | vedclass

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

Question

(A) The activity $R$ of a radioactive sample is given by the radioactive decay law: $R = R_{0} e^{-\lambda t}$,where $R_{0}$ is the initial activity a

Vedclass · Accepted Answer

$R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}$

Vedclass · Answer

(A) The activity $R$ of a radioactive sample is given by the radioactive decay law: $R = R_{0} e^{-\lambda t}$,where $R_{0}$ is the initial activity at $t=0$.At time $t_{1}$,the activity is $R_{1} = R_{0} e^{-\lambda t_{1}}$.At time $t_{2}$,the activity is $R_{2} = R_{0} e^{-\lambda t_{2}}$.Dividing the expression for $R_{1}$ by $R_{2}$,we get:$\frac{R_{1}}{R_{2}} = \frac{R_{0} e^{-\lambda t_{1}}}{R_{0} e^{-\lambda t_{2}}} = e^{-\lambda t_{1} - (-\lambda t_{2})} = e^{-\lambda(t_{1}-t_{2})}$.Therefore,$R_{1} = R_{2} e^{-\lambda(t_{1}-t_{2})}$.

In a radioactive material,the activity at time $t_{1}$ is $R_{1}$ and at a later time $t_{2}$,it is $R_{2}$. If the decay constant of the material is $\lambda$,then:

Similar Questions

$A$ radioactive nucleus $A$ with a half-life $T$ decays into a nucleus $B$. At $t = 0$,there is no nucleus $B$. At some time $t$,the ratio of the number of $B$ to that of $A$ is $0.3$. Then,$t$ is given by

$A$ radioactive element emits $200$ particles per second. After $3$ hours,$25$ particles per second are emitted. The half-life period of the element will be .......... $minutes$.

$A$ radioactive material of half-life $2.5$ hours emits radiation that is $32$ times the safe maximum level. The time (in hours) after which the material can be handled safely is

At time $t = 0$,$N_1$ nuclei of decay constant $\lambda_1$ and $N_2$ nuclei of decay constant $\lambda_2$ are mixed. The decay rate of the mixture is:

Give the different units of radioactivity and define them.

Vedclass Test Series

Exam Paper Generator

Online Exam Module