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After 32 days, 5 milligrams of an 80-milligram sample of a radioactive isotope remains unchanged. What is the half-life of this element?

(1) 8 days (3) 16 days

(2) 2 days (4) 4 days

Verified Answer :

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\(m=m_0 \times (\frac{1}{2})^\frac{t}{t_{1/2}}\)
m - the mass that remains unchanged, m₀ - the inital mass, t - the time of decay, t1⁄2 - the half-life

\(t=32 \ days \\ m=5 \ mg \\ m_0 = 80 \ mg \\ \\ 5 = 80 \times (\frac{1}{2})^\frac{32}{t_{1/2}} \ \ \ \ \ \ \ |\div 80 \\ \frac{5}{80}= (\frac{1}{2})^\frac{32}{t_{1/2}} \\ \frac{1}{16}=(\frac{1}{2})^\frac{32}{t_{1/2}} \\ (\frac{1}{2})^4=(\frac{1}{2})^\frac{32}{t_{1/2}} \\ 4=\frac{32}{t_{1/2}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\times t_{1/2} \\ 4t_{1/2}=32 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |\div 4 \\ t_{1/2}=8\)

The half-life is (1) 8 days.

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