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Hint- This question asks us the brief explanation about the radioactivity. As we can see that in case one Rate of decay of a nucleus is directly proportional to its quantity as we keep the factors, we can find our case one, where else in case two we have the half-life of the substance is 5 days by use. As the substance gets disintegrated by 31/32th by using equation two from above we keep the required factors and clear the case three.
Complete step-by-step answer:
Radioactivity, properties demonstrated by certain types of matter of releasing energy and subatomic particles spontaneously. It is, in essence, a characteristic of individual atomic nuclei. An unstable nucleus will decompose spontaneously, or decay, into a more stable configuration but will do so only in a few specific ways by releasing certain particles or certain types of electromagnetic energy. Radioactive decay is a feature of several elements that occur naturally as well as of isotopes of the elements artificially produced. The rate at which a radioactive element decays through its half-life; i.e. the time taken to decay for half of any given quantity of the isotope.
Step one-
Rate of decay of a nucleus is directly proportional to its quantity i.e. \[\dfrac{{ - dn}}{{dt}}\alpha N\]
Thus, we get \[\dfrac{{dN}}{{dt}} = - \lambda N\] where \[\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\]proportionality is constant.
By integrating both sides we get. \[\int_N^N {\dfrac{{dN}}{N} = \int_0^t { - \lambda t} } \] Where No is the amount of nucleus initially.
\[
\dfrac{N}{{{N_o}}} = - \lambda t \\
\dfrac{N}{{{N_o}}} = {e^{ - \lambda t}} \\
\]
Number of nuclei left after time t, N=\[{N_o}{e^{ - \lambda t}}\]
Step two-
Given above in the question that half-life of a certain substance \[{T_{\dfrac{1}{2}}}\]= 5 days
As the substance gets disintegrated by 31/32th of its initial mass in time t.
Amount of substance left, N = \[\left( {1 - \dfrac{{31}}{{32}}} \right){N_o} = \dfrac{1}{{32}}{N_o}\]
\[
\Rightarrow \dfrac{1}{{32}}{N_o} = {N_o}{e^{ - \lambda t}} \\
\\
\Rightarrow \dfrac{1}{{32}} = {e^{\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}}} \\
\\
\Rightarrow \dfrac{1}{{32}} = - \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} \\
\\
\Rightarrow - 3.4657 = - \dfrac{{0.693}}{5} \times t \\
\]
Hence time t = 25 days.
Note- Radioactivity is the term used to describe the natural process through which certain atoms disintegrate spontaneously, emitting particles and energy as they become different and more stable atoms. This process , also known as radioactive decay, takes place because unstable isotopes tend to become more stable. Radioactivity is measured per unit time in terms of breakdown or decay. Becquerel, equivalent to one decline per second, and the curie, equals 37 billion declines per second, are specific units of radioactivity. If during the nuclear reaction the radioactive isotope is generated, it is called artificial radiation.
Complete step-by-step answer:
Radioactivity, properties demonstrated by certain types of matter of releasing energy and subatomic particles spontaneously. It is, in essence, a characteristic of individual atomic nuclei. An unstable nucleus will decompose spontaneously, or decay, into a more stable configuration but will do so only in a few specific ways by releasing certain particles or certain types of electromagnetic energy. Radioactive decay is a feature of several elements that occur naturally as well as of isotopes of the elements artificially produced. The rate at which a radioactive element decays through its half-life; i.e. the time taken to decay for half of any given quantity of the isotope.
Step one-
Rate of decay of a nucleus is directly proportional to its quantity i.e. \[\dfrac{{ - dn}}{{dt}}\alpha N\]
Thus, we get \[\dfrac{{dN}}{{dt}} = - \lambda N\] where \[\lambda = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\]proportionality is constant.
By integrating both sides we get. \[\int_N^N {\dfrac{{dN}}{N} = \int_0^t { - \lambda t} } \] Where No is the amount of nucleus initially.
\[
\dfrac{N}{{{N_o}}} = - \lambda t \\
\dfrac{N}{{{N_o}}} = {e^{ - \lambda t}} \\
\]
Number of nuclei left after time t, N=\[{N_o}{e^{ - \lambda t}}\]
Step two-
Given above in the question that half-life of a certain substance \[{T_{\dfrac{1}{2}}}\]= 5 days
As the substance gets disintegrated by 31/32th of its initial mass in time t.
Amount of substance left, N = \[\left( {1 - \dfrac{{31}}{{32}}} \right){N_o} = \dfrac{1}{{32}}{N_o}\]
\[
\Rightarrow \dfrac{1}{{32}}{N_o} = {N_o}{e^{ - \lambda t}} \\
\\
\Rightarrow \dfrac{1}{{32}} = {e^{\dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}}} \\
\\
\Rightarrow \dfrac{1}{{32}} = - \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}} \\
\\
\Rightarrow - 3.4657 = - \dfrac{{0.693}}{5} \times t \\
\]
Hence time t = 25 days.
Note- Radioactivity is the term used to describe the natural process through which certain atoms disintegrate spontaneously, emitting particles and energy as they become different and more stable atoms. This process , also known as radioactive decay, takes place because unstable isotopes tend to become more stable. Radioactivity is measured per unit time in terms of breakdown or decay. Becquerel, equivalent to one decline per second, and the curie, equals 37 billion declines per second, are specific units of radioactivity. If during the nuclear reaction the radioactive isotope is generated, it is called artificial radiation.
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