Answer:
It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.
Explanation:
Radioactive decays/reactions always follow a first order reaction dynamic.
Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A
The general expression for rate of reaction for a first order reaction is
(dA/dt) = -kA (Minus sign because it's a rate of reduction)
k = rate constant
(dA/dt) = -kA
(dA/A) = -kdt
∫ (dA/A) = -k ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.
We get
In (A/A₀) = -kt
(A/A₀) = e⁻ᵏᵗ
A(t) = A₀ e⁻ᵏᵗ
Although, we can obtain k from the information on half life.
For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus
T(1/2) = (In2)/k
T(1/2) = 5730 years
k = (In 2)/5730 = 0.000120968 = 0.000121 /year.
So, the amount of C-14 atoms left at any time is given as
A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ
How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?
When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.
Hence,
A(t) = 75%
A₀ = 100%
100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ
e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75
In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768
-0.000121t = -0.28768
t = (0.28768/0.000121) = 2,377.54 = 2378 years
Hope this Helps!!!
