Question

Phosphorous- \[32\] has a half-life of \[14\] days. Starting with \[6.00{\text{ }}g\] of \[^{32}P\] , how many grams will remain after \[56\] days?

Answer 1

Hint: Isotopes are those species which are having atomic numbers but having different atomic masses. The radioisotopes or 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 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.
Suppose, Phosphorous- \[32\] is having \[14.0\] days of half-life. Therefore, it denotes that every \[14.0\] days the sample of Phosphorus- \[32\] gets halved.
The decay follows the half-order kinetics.
\[{t_{1/2}} = \dfrac{{\ln 2}}{k}\]
The given values are;
Half-life of Phosphorous-\[32\] = \[14.0\] days
So, we need to find the Amount remains after \[56\] days.
\[14 = \dfrac{{\ln 2}}{k}\]
\[k = \dfrac{{\ln 2}}{{14}}\]
Let’s write the first order of kinetics;
\[\ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,kt\]
\[\ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\dfrac{{\ln 2}}{{14}} \cdot 56 = 4\ln 2 = \ln {2^4}\]
\[\ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \ln \,16\]
\[\left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,16\]
Now, as the concentration is greater the mass will be greater;
\[C\,\alpha \,M\]
\[\left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\left[ {\dfrac{{{m_0}}}{{{m_t}}}} \right]\, = \,16\]
Total amount of sample = \[6.00{\text{ }}g\]
\[{m_0}\, = \,6.00\,g\]
\[\,\left[ {\dfrac{{6.0}}{{{m_t}}}} \right] = \,16\]
\[{m_t}\, = \,0.375\,g\]
So, the sample of phosphorus- \[32\] will remain \[0.375\,g\] at the end of \[56\] days
Therefore, the answer is \[0.375\,g\] of phosphorus- \[32\] will remain.

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 nuclei which don’t have enough binding energy to hold the nucleus together. phosphorous- \[32\] is a radioactive isotope of phosphorus. It is used for the tracers for cancer detection, radiotherapy for some cancer treatments and tracers for fertilizer uptake analysis.