1) The half-life is the time required for a substance to reduce to half its initial value. In formulas:
[tex] \frac{m(t)}{m_0} = ( \frac{1}{2} )^{t/t_{1/2}}[/tex] (1)
where
m(t) is the amount of substance left at time t
m0 is the initial mass
[tex]t_{1/2}[/tex] is the half-life
In this problem, the half-life of the substance is 20 years:
[tex]t_{1/2} = 20 y[/tex]
therefore, the fraction of sample left after t=40 years will be
[tex] \frac{m(t)}{m_0}=( \frac{1}{2})^ \frac{40 y}{20 y} = ( \frac{1}{2})^2 = \frac{1}{4} [/tex]
So, only 1/4 of the original sample will be left, which corresponds to 25%.
2) We can use again formula (1), by re-arranging it:
[tex]m_0 = \frac{m(t)} {( \frac{1}{2} )^{ \frac{t}{t_{1/2} }}}[/tex]
If we use m(t)=10 g (mass of uranium left at time t), and [tex]t=4 t_{1/2}[/tex] (the time is equal to 4 half lifes), we get
[tex]m_0 = \frac{10 g}{ (\frac{1}{2})^4 } =16 \cdot 10 g = 160 g[/tex]
So, the initial sample of uranium was 160 g.
