Question

Phosphorus- \[32\] is radioactive and has a half-life of \[14.3\] days. How long would it take a sample to decay from \[2.20{\text{ }}mg\] to \[600\,\mu g\] ?

Answer 1

Hint:Isotopes are those species which have atomic numbers but have different atomic masses. The radioisotopes of the radioactive isotopes are named as a radionuclide or radioactive nuclide. These species are those, whose atomic number is the same but differs with masses whose nuclei are unstable and vanishes the excess energy by spontaneously emitting radiation in the form of alpha, beta and gamma rays.

Complete step-by-step answer:Half-life \[(\,{t_{1/2}}\,)\] is the time required for a quantity to reduce into half of its initial value of a compound.
The nuclear half-life conveys about how much time passes in the order for half of the atoms present in the initial sample to undergo radioactive decay. Due to this, the half-life is important to know. Half-life says about the time intervals that are expected from radioactive isotopes to be halved.
The equation for the first-order decay;
\[{A_t}\, = \,{A_0}{e^{ - kt}}\]
Suppose, Phosphorus- \[32\] is having \[14.3\] days of half-life. Therefore, it denotes that every \[14.3\] days the sample of Phosphorus- \[32\] gets halved.
The decay actually follows the half-order kinetics.
\[{t_{1/2}} = \dfrac{{\ln 2}}{k}\]
The given values are;
Half-life of phosphorus- \[32\] \[ = \,14.3\] days
So, we need to find the days to decay from \[2.20{\text{ }}mg\] to \[600\,\mu g\].
\[14.3\, = \,\dfrac{{\ln 2}}{k}\]
\[\Rightarrow k\, = \,\dfrac{{\ln 2}}{{14.3}}\]
\[\Rightarrow k\, = \,\dfrac{{0.693}}{{14.3}}\]
\[\Rightarrow k\, = \,0.048\,{d^{ - 1}}\]
Let’s re-write the first order of kinetics;
\[{A_t}\, = \,{A_0}{e^{ - kt}}\]
Applying natural logs on both the side;
\[\ln \,{A_t}\, = \,\ln \,{A_0}\, - \,kt\]
Where,
\[{A_t}\, = \] The amount of radioisotope at \[t\] time
\[{A_0}\, = \] Initial amount of radioisotope
\[k\, = \] Rate constant
\[t\, = \] Time
The given values are;
\[{A_t}\, = \,600\,\mu g\, = \,0.60\,mg\]
\[{A_0}\, = \,2.20{\text{ }}mg\]
\[k\, = \,0.048\,{d^{ - 1}}\]
\[t\, = \,?\]
Let’s substitute the values in above equation;
\[\ln \,0.60\, = \,\ln \,2.2\, - \,0.048t\]
\[0.048t\, = \,\ln \,2.2\, - \,\ln \,0.60\]
\[\ln \left[ {\dfrac{{2.2}}{{0.6}}} \right] = \,0.048t\]
\[\ln \,[3.6]\, = \,0.048t\]
\[1.28\, = \,0.048t\]
\[t\, = \,\dfrac{{1.28}}{{0.048}}\]
\[t\, = \,26.68\,days\]
The phosphorus- \[32\] would it take \[26.68\,\] days to decay from \[2.20{\text{ }}mg\] to \[600\,\mu g\]

Note:Radioactive decay is the spontaneous process of breakdown of an atomic nucleus which results in the releasing of energy and the matter from the nucleus. The radioisotopes have the unstable nucleus which doesn’t have enough binding energy to hold the nucleus together.