A piece of wood from a recently cut tree show | vedclass

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

Question

(C) Given: Initial activity $A_0 = 20$ decays/min.Final activity $A = 2$ decays/min.Half-life $T_{1/2} = 5730$ years.We know that the activity $A$ at

Vedclass · Accepted Answer

$19039$

Vedclass · Answer

(C) Given: Initial activity $A_0 = 20$ decays/min.Final activity $A = 2$ decays/min.Half-life $T_{1/2} = 5730$ years.We know that the activity $A$ at time $t$ is given by $A = A_0 e^{-\lambda t}$,where $\lambda = \frac{0.693}{T_{1/2}}$.Rearranging for $t$,we get $t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A}ight) = \frac{T_{1/2}}{0.693} 	imes 2.303 	imes \log_{10}\left(\frac{A_0}{A}ight)$.Substituting the values: $t = \frac{5730}{0.693} 	imes 2.303 	imes \log_{10}\left(\frac{20}{2}ight)$.Since $\log_{10}(10) = 1$,we have $t = \frac{5730 	imes 2.303}{0.693} \approx 5730 	imes 3.322$.$t \approx 19039$ years.

Similar Questions

The half-life of a substance is $20 \ min$. The time taken between $33\%$ decay and $67\%$ decay will be ....... $min$.

If the radioactive decay constant of radium is $1.07 \times 10^{-4}$ per year,then its half-life period is approximately equal to ......... $years$.

$A$ fresh radioactive sample is given at $t = 0$. Its decay fraction is $\frac{1}{5}$ at $t_1$ instant and $\frac{4}{5}$ at $t_2$ instant. Its mean life is

Decay constants of two radioactive samples $A$ and $B$ are $15x$ and $3x$ respectively. They have an equal number of initial nuclei. The ratio of the number of nuclei left in $A$ and $B$ after a time $t = \frac{1}{6x}$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module