Respuesta :
Answer
a) The half-life of the substance is less than 200 days.
b) A formula that models the amount A of the substance in the table after x days.
A(x) = 3 e⁻⁰.⁰⁰⁶⁷³³ˣ
A(x) = 3 e^(-0.006733x)
c) Half life = 103 days
Explanation
a) The half life of a substance is defined as the time taken for the substance to reduce to half of its original mass.
For this question, since this substance starts with a mass of 3 milligrams, we can see that the sbstance reaches half of its mass (1.5 milligrams) way before 200 days, (in fact, it has to reach half of its mass a few days after 100 days).
So, the half life of this substance is less than 200 days.
b) Radioactive decays are described using first order decay dynamics which is given by the general formula
A = A₀ e⁻ᵏˣ
where
A = Amount of the substance at any time x
A₀ = Initial amount of the substance = 3 milligrams
k = decay constant or rate constant for the decay process
x = time, measured in days for this question
So, for this question, we will insert some of the values of A at different times to obtain the unknowns
At x = 100 days, A = 1.53 milligrams
A = A₀ e⁻ᵏˣ
1.53 = 3 e^(-k × 100)
1.53 = 3 e^(-100k)
When we divide both sides by 3, we get
e⁻¹⁰⁰ᵏ = (1.53/3)
e⁻¹⁰⁰ᵏ = 0.51
Taking the natural logarithms of both sides
In e⁻¹⁰⁰ᵏ = In 0.51
-100k = In 0.61
-100k = -0.6733
Divide both sides by -100
k = 0.006733
k = (6.733 × 10⁻³)
If we use any of the two other corresponding sets of values, we will obtain the same value for k
At x = 200 days, A = 0.740 milligrams
At x = 300 days, A = 0.367 milligrams
So, we can write the formula that models the amount A of the substance that is left after x days now
A(x) = A₀ e⁻ᵏˣ
A₀ = 3
k = 0.006733 = (6.733 × 10⁻³)
A(x) = 3 e⁻⁰.⁰⁰⁶⁷³³ˣ
A(x) = 3 e^(-0.006733x)
c) For the half life, the half life of a decaying substance due to radioactivity is given as
T = (In 2)/k
In 2 = 0.6931
k = decay constant calculated above = 0.006733
T = half-life = ?
T = (0.6931/0.006733) = 103 days to the nearest whole number.
Hope this Helps!!!
