Answer:
2.576 mg of gold-198 remains after 3 days
Explanation:
N = No(0.5)^t/t1/2
No is the initial amount of gold-198 = 5.6 mg
t is the time taken for gold-198 to reduce to a certain amount (N) = 3 days
t1/2 is the half-life of gold-198 = 2.7 days
N = 5.6(0.5)^3/2.7 = 5.6(0.5)^1.11 = 5.6×0.46 = 2.576 mg
Answer:
Nt = 2.59 mg
The amount of gold-198 remaining after 3.0 days is 2.59 mg
Explanation:
Decay of a substance can be expressed mathematically as
Nt = No.e^(-λt) .....1
Where
Nt = amount remaining at time t
No = initial amount = 5.6 mg
t = decay time = 3 days
λ = decay constant
And decay constant can be expressed in terms of half life as;
λ = ln(2)/tₕ ....2
Where;
tₕ = half life = 2.7 days
Substituting equation 2 to 1.
Nt = No.e^(-t × ln(2)/tₕ)
Substituting the values, we have;
Nt = 5.6.e^(-3 × ln(2)/2.7)
Nt = 2.59 mg
The amount of gold-198 remaining after 3.0 days is 2.59 mg
