Question

An 80.0 g sample of iodine-131 was placed in a sealed vessel forty days ago. Only 2.5 g of this isotope is now left. What is its half-life?

Answer 1

$m=m_0 \times (\frac{1}{2})^\frac{t}{T}$
m - the mass that remains, m₀ - the initial mass, t - time, T - the half-life

$m_0=80 \ g \\ m=2.5 \ g \\ t=40 \ d \\ \\ 2.5=80 \times (\frac{1}{2})^\frac{40}{T} \\ \frac{2.5}{80}=(\frac{1}{2})^\frac{40}{T} \\ \frac{1}{32}=(\frac{1}{2})^\frac{40}{T} \\ (\frac{1}{2})^5=(\frac{1}{2})^\frac{40}{T} \\ 5=\frac{40}{T} \\ 5T=40 \\ T=\frac{40}{5} \\ T=8$

The half-life is 8 days.

Answer 2

Answer:

The answer is 8 days