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Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days. If you begin with 48.0 mg of this isotope, what mass remains after 47.9 days have passed?

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Answer:

After 47.9 days, will remain 14.5mg of the isotope

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:  

Ln[A] = -Kt + ln[A]₀

Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration.

We can find rate constant from half-life as follows:

[tex]t_{1/2} = \frac{ln2}{K}[/tex]

K = ln 2 / 27.7 days

K = 0.025 days⁻Âč

Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻Âč and t = 47.9 days:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.025 days⁻Âč*47.9 days + ln (48.0mg)

ln [A] = 2.6726

[A] = e^ (2.6726)

[A] = 14.5mg

After 47.9 days, will remain 14.5mg of the isotope

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