(A) Initial activity $R_0 = 1 \, \mu Ci = 10^{-6} \times 3.7 \times 10^{10} \, dps = 3.7 \times 10^4 \, dps$.Activity of $1 \, cm^3$ sample at $t = 5

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(A) Initial activity $R_0 = 1 \, \mu Ci = 10^{-6} \times 3.7 \times 10^{10} \, dps = 3.7 \times 10^4 \, dps$.Activity of $1 \, cm^3$ sample at $t = 5 \, hours$ is $r = \frac{296}{60} \, dps \approx 4.933 \, dps$.The activity of the total blood volume $V$ at time $t$ is given by $R = R_0 e^{-\lambda t}$.Since the radioactive substance is uniformly mixed,the activity per unit volume is $r = \frac{R}{V}$.Therefore,$V = \frac{R}{r} = \frac{R_0 e^{-\lambda t}}{r}$.Substituting the values: $V = \frac{3.7 \times 10^4 \times 0.7927}{4.933} \approx 5945 \, cm^3$.Since $1000 \, cm^3 = 1 \, L$,the total volume $V \approx 5.94 \, L$.

Class 12 Physics Nuclei Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

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$A$ small quantity of solution containing $Na^{24}$ radionuclide of activity $1 \, \mu Ci$ is injected into the blood of a person. $A$ sample of the blood of volume $1 \, cm^3$ taken after $5 \, hours$ shows an activity of $296$ disintegrations per minute. What will be the total volume of the blood in the body of the person? Assume that the radioactive solution mixes uniformly in the blood of the person: ............ $L$ (Take $1 \, Ci = 3.7 \times 10^{10}$ disintegrations per second and $e^{-\lambda t} = 0.7927$; where $\lambda$ is the disintegration constant).

A
$5.94$
B
$2$
C
$317$
D
$1$

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Nuclei

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Law of Radioactivity by Rutherford and Soddy and Half Life and Mean Life

$A$ radioactive nuclide is produced at a constant rate of $n$ per second (e.g.,by bombarding a target with neutrons). If the number of nuclei at $t = 0$ is $N_0$,the number of nuclei $N$ at time $t$ is given by (where $\lambda$ is the decay constant):

DifficultAIIMS 2014

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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 .......... $years$.

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The fossil bone has a ${}^{14}C:{}^{12}C$ ratio,which is $\frac{1}{16}$ of that in a living animal bone. If the half-life of ${}^{14}C$ is $5730 \, years$,then the age of the fossil bone is .......... $years$.

MediumAIIMS 2006

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$A$ radioactive element decays to form a stable nuclide. The graph representing the number of radioactive nuclei $(N)$ versus time $(t)$ is:

MediumAIIMS 2012

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Following statements related to radioactivity are given below:
$(A)$ Radioactivity is a random and spontaneous process and is dependent on physical and chemical conditions.
$(B)$ The number of undecayed nuclei in the radioactive sample decays exponentially with time.
$(C)$ Slope of the graph of $\log_{e}$ (no. of undecayed nuclei) $Vs.$ time represents the negative reciprocal of mean life time $(-\frac{1}{\tau})$.
$(D)$ Product of decay constant $(\lambda)$ and half-life time $(T_{1/2})$ is constant,equal to $\ln(2)$.
Choose the most appropriate answer from the options given below.

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$A$ radioactive nuclide is produced at a constant rate of $n$ per second (e.g.,by bombarding a target with neutrons). If the number of nuclei at $t = 0$ is $N_0$,the number of nuclei $N$ at time $t$ is given by (where $\lambda$ is the decay constant):

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 fossil bone has a ${}^{14}C:{}^{12}C$ ratio,which is $\frac{1}{16}$ of that in a living animal bone. If the half-life of ${}^{14}C$ is $5730 \, years$,then the age of the fossil bone is .......... $years$.

$A$ radioactive element decays to form a stable nuclide. The graph representing the number of radioactive nuclei $(N)$ versus time $(t)$ is:

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