Two radioactive materials A and B have decay | vedclass

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

Question

Given : $\lambda_{A}=5 \lambda$, $\lambda_{B}=\lambda$

At $t=0,\left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}$

$\frac{N_{A}}{N_{B}}=\left(\frac{1

Vedclass · Accepted Answer

$\frac{1}{{2\lambda }}$

Vedclass · Answer

Given : $\lambda_{A}=5 \lambda$, $\lambda_{B}=\lambda$

At $t=0,\left(N_{0}ight)_{A}=\left(N_{0}ight)_{B}$

$\frac{N_{A}}{N_{B}}=\left(\frac{1}{e}ight)^{2}$

According to radioactive decay, $\frac{N}{N_{0}}=e^{-\lambda t}$

$	herefore \,\frac{N_{A}}{\left(N_{0}ight)_{A}} =e^{-\lambda_{A} t} $  ..... $(i)$

$\frac{N_{B}}{\left(N_{0}ight)_{B}} =e^{-\lambda_{B} t}$  ..... $(ii)$

Divide $(i)$ by $(ii)$, we get

$\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}ight) t}$  or, $\frac{N_{A}}{N_{B}}=e^{-(5 \lambda-\lambda) t}$

or, $\left(\frac{1}{e}ight)^{2}=e^{-4 \lambda t}$ or,  $\left(\frac{1}{e}ight)^{2}=\left(\frac{1}{e}ight)^{4 \lambda t}$

or, $4 \lambda t=2$ $ \Rightarrow \quad t=\frac{2}{4 \lambda}=\frac{1}{2 \lambda}$

Two radioactive materials $A$ and $B$ have decay constant $5\lambda$ and $\lambda$ respectively.At $t=0$ they have the same number of nuclei, then the ratio of the number of nuclei of $A$ to that $B$ will be $(1/e)^2$ after a time interval

Similar Questions

The curve between the activity $A$ of a radioactive sample and the number of active atoms $N$ is

Half lives of two radioactive nuclei $A$ and $B$ are $10\, minutes$ and $20\, minutes$, respectively. If, initially a sample has equal number of nuclei, then after $60$ $minutes$ , the ratio of decayed numbers of nuclei $A$ and $B$ will be

The decay constant of a radioactive element is $1.5 \times {10^{ - 9}}$ per second. Its mean life in seconds will be

Consider an initially pure $M$ gm sample of$_ A{X}$, an isotope that has a half life of $T$ hour, what is it’s initial decay rate ($N_A$ = Avogrado No.)