1) The half-life is the time required for a substance to reduce to half its initial value. In formulas:
![(m(t))/(m_0) = ( (1)/(2) )^{t/t_(1/2)}](https://img.qammunity.org/2019/formulas/physics/college/15ppe1dz4bxye96culp31ygxvlwfe8hie1.png)
(1)
where
m(t) is the amount of substance left at time t
m0 is the initial mass
![t_(1/2)](https://img.qammunity.org/2019/formulas/physics/college/2fqvmki85nx36yspopvaz3mlq36cn81lba.png)
is the half-life
In this problem, the half-life of the substance is 20 years:
![t_(1/2) = 20 y](https://img.qammunity.org/2019/formulas/physics/college/k754zx7tvjw5gakmd01dv76jwk8g8estj4.png)
therefore, the fraction of sample left after t=40 years will be
![(m(t))/(m_0)=( (1)/(2))^ (40 y)/(20 y) = ( (1)/(2))^2 = (1)/(4)](https://img.qammunity.org/2019/formulas/physics/college/1enclwmq8ppqyjlbt84pmca1smyasipra1.png)
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:
![m_0 = \frac{m(t)} {( (1)/(2) )^{ (t)/(t_(1/2) )}}](https://img.qammunity.org/2019/formulas/physics/college/uxevj8515ogb4qhuyc7ju6ecw8ozi8ld4y.png)
If we use m(t)=10 g (mass of uranium left at time t), and
![t=4 t_(1/2)](https://img.qammunity.org/2019/formulas/physics/college/5aux1kw1mtpuue9wenyy0l0dtxsgfqydvc.png)
(the time is equal to 4 half lifes), we get
![m_0 = (10 g)/( ((1)/(2))^4 ) =16 \cdot 10 g = 160 g](https://img.qammunity.org/2019/formulas/physics/college/nch9wr42qwirvybyyvto17k5qh278s6pyc.png)
So, the initial sample of uranium was 160 g.