The activity of a radioactive material is 6.4 | vedclass

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

Question

$A_{0}=6.4 	imes 10^{-4}$ Curie

$T _{1 / 2}=5 	ext { days }=\frac{\ln 2}{\lambda}$

$A = A _{0} e ^{-\lambda t }$

$5 	imes 10^{-6}=6.4 	im

Vedclass · Accepted Answer

$35$

Vedclass · Answer

$A_{0}=6.4 	imes 10^{-4}$ Curie

$T _{1 / 2}=5 	ext { days }=\frac{\ln 2}{\lambda}$

$A = A _{0} e ^{-\lambda t }$

$5 	imes 10^{-6}=6.4 	imes 10^{-4} e ^{-\lambda t }$

$\frac{5}{6.4} 	imes 10^{-2}=e^{-\lambda t}$

$7.8 	imes 10^{-3}= e ^{-\lambda t}$

$\log \left(7.8 	imes 10^{-3}ight)=-\lambda t \ln e$

$\ln \left(7.8 	imes 10^{-3}ight)=-\frac{\lambda n 2}{5} \cdot t$

$	herefore \frac{5 	imes 4.853}{0.693}= t =35 	ext { days }$

The activity of a radioactive material is $6.4 \times 10^{-4}$ curie. Its half life is $5\; days$. The activity will become $5 \times 10^{-6}$ curie after $.......day$

Similar Questions

The half life period of a radioactive substance is 60 days. The time taken for $\frac{7}{8}$ th of its original mass to disintegrate will be $......days$

Decay constant of radium is $\lambda $. By a suitable process its compound radium bromide is obtained. The decay constant of radium bromide will be

$A$ and $B$ are two radioactive substances whose half lives are $1$ and $2$ years respectively. Initially $10\, g$ of $A$ and $1\,g$ of $B$ is taken. The time (approximate) after which they will have same quantity remaining is ........... $years$

A radioactive material has a half life of $10$ days. What fraction of the material would remain after $30$ days

$N$ atoms of a radioactive element emit $n$ alpha particles per second. The half life of the element is