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.

Answer:

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.