
The half-life of $ {}^{215}At $ is $ 100\mu s $ .The time taken for the radioactivity of a sample of $ {}^{215}At $ to decay $ \dfrac{1}{{16}}th $ of initial value is
$ \left( A \right)400\mu s \\
\left( B \right)6.3\mu s \\
\left( C \right)40\mu s \\
\left( D \right)300\mu s \\ $
Answer
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Hint: In order to solve this question, we are going to first determine the decay constant from the half-life of $ {}^{215}At $ as given in the question. After that, the time for the concentration of the $ {}^{215}At $ to decay $ \dfrac{1}{{16}}th $ of initial value is calculated by putting the values in the law of radioactive decay equation.
According to the law of radioactive decay, the concentration of $ {}^{215}At $ at a time $ t $ is given by
$ N\left( t \right) = {N_0}{e^{ - \lambda t}} $
The half-life of the radioactive element is given by
$ {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } $
Complete step by step solution:
According to the law of radioactive decay, the concentration of $ {}^{215}At $ at a time $ t $ is given by
$ N\left( t \right) = {N_0}{e^{ - \lambda t}} $
Now as we know that the half-life of the radioactive element, i.e. in which the concentration of $ {}^{215}At $ remains half of its initial value is given by
$ {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } $
Now putting the value of the decay constant, $ \lambda = \dfrac{{\ln 2}}{{{t_{\dfrac{1}{2}}}}} $ in the above equation for the concentration measurement
Putting these values for the decay to the $ \dfrac{1}{{16}}th $ of initial value:
$ \dfrac{1}{{16}} = {e^{ - \left( {\dfrac{{0.693}}{{100}}} \right)t}} \\
\Rightarrow \ln \left( {\dfrac{1}{{16}}} \right) = - \dfrac{{0.693}}{{100}}t \\
\Rightarrow t = 400\mu s \\ $
Thus, the option $ \left( A \right)400\mu s $ is the correct answer.
Note:
The probability per unit time that a nucleus will decay is constant, independent of time. The decay constant is represented by $ \lambda $ . The number of radioactive elements undergoing decay per unit time, is proportional to the total number of nuclei in the sample material. Half – life of the radioactive element is the time in which the concentration of the radioactive element remains half of its initial concentration.
According to the law of radioactive decay, the concentration of $ {}^{215}At $ at a time $ t $ is given by
$ N\left( t \right) = {N_0}{e^{ - \lambda t}} $
The half-life of the radioactive element is given by
$ {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } $
Complete step by step solution:
According to the law of radioactive decay, the concentration of $ {}^{215}At $ at a time $ t $ is given by
$ N\left( t \right) = {N_0}{e^{ - \lambda t}} $
Now as we know that the half-life of the radioactive element, i.e. in which the concentration of $ {}^{215}At $ remains half of its initial value is given by
$ {t_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda } $
Now putting the value of the decay constant, $ \lambda = \dfrac{{\ln 2}}{{{t_{\dfrac{1}{2}}}}} $ in the above equation for the concentration measurement
Putting these values for the decay to the $ \dfrac{1}{{16}}th $ of initial value:
$ \dfrac{1}{{16}} = {e^{ - \left( {\dfrac{{0.693}}{{100}}} \right)t}} \\
\Rightarrow \ln \left( {\dfrac{1}{{16}}} \right) = - \dfrac{{0.693}}{{100}}t \\
\Rightarrow t = 400\mu s \\ $
Thus, the option $ \left( A \right)400\mu s $ is the correct answer.
Note:
The probability per unit time that a nucleus will decay is constant, independent of time. The decay constant is represented by $ \lambda $ . The number of radioactive elements undergoing decay per unit time, is proportional to the total number of nuclei in the sample material. Half – life of the radioactive element is the time in which the concentration of the radioactive element remains half of its initial concentration.
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