Iron‑59 is used to study iron metabolism in the spleen. Its half‑life is 44 days. How many days would it take a 28.0 g sample of iron‑59 to decay to 0.875 g?

Any radioactive isotope is tend to decay with time. So the rate of decay of the radioactive isotopes is termed as disintegration constant. Since, the initial mass of the radioactive isotope is given along with the reducing mass. In order to determine the time required to reduce the mass of the radioisotope from 28 g to 0.875 g, first the disintegration constant is need to be determined. The disintegration constant can be obtained from half life time of the isotope. As half life time is the measure of time required to reduce half of the concentration of the isotope.

Half life time = 0.6932/disintegration constant

44 = 0.6932/λ

λ = 0.6932/44=0.0158

So, with this values of disintegration constant, initial mass and final mass, the time required to reduce from initial to final mass can be obtained using law of disintegration constant as follows.

N = Noe^(-λt)

[tex]t= -\frac{1}{disintegration constant}* ln(\frac{N}{N_{0} } )\\ \\t = -\frac{1}{0.0158}*ln(\frac{0.875}{28})\\ \\t = -\frac{1}{0.0158}*ln(0.03125)\\\\t = -63.29*(-3.466)\\\\t=219 days.[/tex]

Thus, the radioisotope will take 219 days to decay from 28 g to 0.875 g.

raphealnwobi raphealnwobi · Answer 2 · 2021-12-03T14:14:13+01:00

It would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g

The half life of a substance is the amount of time it takes to decay to half of the initial value. It is given by:

[tex]N(t)=N_o(\frac{1}{2}) ^\frac{t}{t_\frac{1}{2} } \\\\N(t)=amount\ after\ t\ years, N_o=original\ value,t_\frac{1}{2} =half\ life\\\\Given\ that:\\\\N(t)=0.875g,N_o=28g,t_\frac{1}{2} =44\ days. hence:\\\\\\0.875=28(\frac{1}{2} )^\frac{t}{44} \\\\t=220\ days[/tex]

Hence it would take 220 days for 28.0 g sample of iron‑59 to decay to 0.875 g

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