• # question_answer The half-life of $_{6}^{14}C$ is 5700 years. What does it mean? Two radioactive nuclei X and Y initially contain equal number of atoms. Their half-life is 1h and 2h, respectively. Calculate the ratio of their rates of disintegration after 2 hours.

When we say that the half-life of $_{6}^{14}C$ is 5700 years, it means that quantity of radioactive material as well as radioactivity of $_{6}^{14}C$ is reduced to half of its original value in 5700 years. In the problem given here, ${{({{N}_{0}})}_{X}}={{({{N}_{0}})}_{Y}}={{N}_{0}}$ (say) So,       ${{({{T}_{1/2}})}_{X}}=1h$ and ${{({{T}_{1/2}})}_{y}}=2h$ $\therefore$      $\frac{{{\lambda }_{X}}}{{{\lambda }_{Y}}}=\frac{{{({{T}_{1}}/2)}_{Y}}}{{{({{T}_{1}}/2)}_{X}}}=\frac{2}{1}=2$ Now, after 2 hours (i.e. 2 half-lives of X),             ${{(N)}_{X}}={{({{N}_{0}})}_{X}}/4={{N}_{0}}/4$ For Y it is only, one half-life, hence ${{(N)}_{Y}}=\frac{{{({{N}_{0}})}_{Y}}}{2}=\frac{{{({{N}_{0}})}_{X}}}{2}=\frac{{{N}_{0}}}{2}$ If activities at t =2 h be ${{R}_{X}}$ and ${{R}_{Y}}$  respectively, then ${{R}_{X}}={{\lambda }_{X}}\cdot {{(N)}_{X}}$ and ${{R}_{\lambda }}={{\lambda }_{Y}}\cdot {{(N)}_{Y}}$ $\therefore$      $\frac{{{R}_{X}}}{{{R}_{Y}}}=\frac{{{\lambda }_{X}}{{(N)}_{X}}}{{{\lambda }_{Y}}{{(N)}_{Y}}}=\frac{2\times ({{N}_{0}}/4)}{{{N}_{0}}/2}=2\times \frac{2}{4}=\frac{1}{1}$ Thus, rate of disintegration, after 2 hours will be same for both nuclides.
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