The graph shows ln left| frac{dN(t)}{dt} righ | vedclass

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

Question

(B) The activity of a radioactive sample is given by $A = \left| \frac{dN(t)}{dt} \right| = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm o

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) The activity of a radioactive sample is given by $A = \left| \frac{dN(t)}{dt} ight| = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln A = \ln(\lambda N_0) - \lambda t$.This is a linear equation of the form $y = mx + c$,where the slope $m = -\lambda$.From the given graph,the slope is $	ext{slope} = \frac{3 - 4}{6 - 4} = \frac{-1}{2}$.Thus,$-\lambda = -\frac{1}{2}$,which gives the decay constant $\lambda = 0.5 	ext{ year}^{-1}$.The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.5} = 1.386 	ext{ years}$.The number of nuclei remaining after time $t$ is $N(t) = N_0 e^{-\lambda t} = N_0 \left( \frac{1}{2} ight)^{t/T_{1/2}}$.Given $t = 4.16 	ext{ years}$,the number of half-lives elapsed is $n = \frac{4.16}{1.386} \approx 3$.Therefore,the remaining fraction is $\frac{N(t)}{N_0} = \left( \frac{1}{2} ight)^3 = \frac{1}{8}$.Since the number of nuclei decreases by a factor of $p$,we have $\frac{N(t)}{N_0} = \frac{1}{p} = \frac{1}{8}$,which implies $p = 8$.

The graph shows $\ln \left| \frac{dN(t)}{dt} \right|$ versus $t$ for a radioactive sample. If the number of radioactive nuclei decreases by a factor of $p$ after $4.16$ years,then $p = $.....

Similar Questions

An archaeologist analyses the wood in a prehistoric structure and finds that the ratio of $C^{14}$ (half-life $= 5700 \, years$) to $C^{12}$ is only one-fourth of that found in the cells of buried plants. The age of the wood is about .......... $years$.

The radioactivity of a sample is $R_1$ at a time $T_1$ and $R_2$ at a time $T_2$. If the half-life of the specimen is $T$,then the number of atoms that have disintegrated in the time interval $(T_2 - T_1)$ is proportional to

The half-life of a radioactive sample is $5 \,s$. If the initial mass of the sample is $60 \,g$, then the time required to reduce the sample to $7.5 \,g$ is (in $\,s$)

The radiation emitted by a radioactive substance with a half-life of $30 \ min$ is measured by a Geiger-Muller counter. If the count rate drops to $5 \ \text{disintegrations/sec}$ in $2 \ \text{hours}$, find the initial disintegration rate.

The average life $T$ and the decay constant $\lambda$ of a radioactive nucleus are related as

Vedclass Test Series

Exam Paper Generator

Online Exam Module