Answer:
(a)y(t)=30exp(-0.0231t)
(b)y(20)=18.9mg
Step-by-step explanation:
At a particular time t, the mass of a radioactive substance like Cesium-137 is governed by the equation:
N=N₀e⁻ᵏᵗ where k=ln 2/half life
(a)Mass that remains after t years
Half Life= 30 years
k= ln2/30=0.0231
Initial Mass, N₀=30mg
Therefore the mass N that remains at time t
N=N₀e⁻ᵏᵗ
N=30exp(-0.0231t)
y(t)=30exp(-0.0231t)
(b)We want to determine how much of the sample remains after 20 years.
At t=20 years
y(t)=30exp(-0.0231t)
y(20)=30exp(-0.0231X20)
=30 X 0.63
y(20)=18.9mg