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# The rate of radioactive disintegration at an instant for a radioactive sample of half life $2.2 \times {10^9}s$ is ${10^{10}}{s^{ - 1}}$ . The number of radioactive atoms in that sample at that instant is:(A) $3.17 \times {10^{19}}$(B) $3.17 \times {10^{20}}$(C) $3.17 \times {10^{17}}$(D) $3.17 \times {10^{18}}$

Last updated date: 15th Jun 2024
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Hint: According to the law of radioactive decay when a radioactive material undergoes decay the number of nuclei that is undergoing the decay per unit time is directly proportional to the total number of nuclei in the given sample. The direct proportionality is removed by the introduction of a constant called the decay constant represented by $\lambda$ .
Half life of a radioactive material is the time required for it to reduce to half of its initial value during decay. The decay constant is related to half time.
Formulas used: We will be using the formula for mathematical representation of the law of radioactive decay, $\dfrac{{\Delta N}}{{\Delta t}} = \lambda N$ where $\dfrac{{\Delta N}}{{\Delta t}}$ is the rate of decay of the radioactive material, $\lambda$ is the decay constant, and $N$ is the total number of radioactive nuclei in the sample.
We will also be using the formula to relate half life of the radioactive nuclei to the half life of the sample, ${t_{1/2}} = \dfrac{{\ln 2}}{\lambda }$ where ${t_{1/2}}$ is the half-life of the sample, $\lambda$ is the decay constant of the sample and $\ln 2 = 0.693$ .

We know that radioactive decay is the process by which an unstable atomic nucleus loses its energy by radiation. The material with unstable nuclei is called radioactive. According to the law of radioactive decay the rate of radioactivity of the material can be directly proportional to the total number of nuclei in the sample.
We know from the problem that the half life of the sample is ${t_{1/2}} = 2.2 \times {10^9}s$.
We know that the half life of a sample is related to the decay constant by, ${t_{1/2}} = \dfrac{{\ln 2}}{\lambda }$ substituting the known values we get, $2.2 \times {10^9}s = \dfrac{{0.693}}{\lambda }$ .
Solving for $\lambda$ we get,
$\lambda = \dfrac{{0.693}}{{2.2 \times {{10}^9}}} = 0.315 \times {10^{ - 9}}{s^{ - 1}}$
Now that we have the decay constant, we know that the rate of decay is given, $R = \dfrac{{\Delta N}}{{\Delta t}} = {10^{10}}{s^{ - 1}}$
We know that $\dfrac{{\Delta N}}{{\Delta t}} = \lambda N$ . Thus, substituting the known values, we get,
${10^{10}}{s^{ - 1}} = \left( {0.315 \times {{10}^{ - 9}}{s^{ - 1}}} \right) \times N$
$\Rightarrow N = \dfrac{{{{10}^{10}}{s^{ - 1}}}}{{0.315 \times {{10}^{ - 9}}{s^{ - 1}}}}$
Solving for $N$ we get,
$\Rightarrow N = 3.174 \times {10^{19}}$

Thus, there will be a total of $N = 3.174 \times {10^{19}}$ nuclei in the sample.

Note: Radium is a radioactive element and was discovered by Madame Marie Curie and hence the older unit of radioactivity was named after her to be curie with the symbol $Ci$ .