Assuming that the original activity of carbon - 14 was 15.3 disintegrations/min/g, what is the present disintegration rate (in disintegrations/min/g) in a sample of parchment that is 2.10×10³ years old?
(t_1/2 = 5730 years)

Question

Quizplus · Accepted Answer

We use the formula for radioactive decay:

N(t) = N0 * (1/2)^(t/t1/2)

Where:
N(t) = the number of nuclei at time t
N0 = the initial number of nuclei
t1/2 = the half-life
t = elapsed time

We can rewrite this as:

N(t) = N0 * e^(-λt)

Where λ is the decay constant, given by:

λ = ln(2)/t1/2

Let's plug in the values we have:

t = 2.10×10^13 years
t1/2 = 5730 years
N0/N(t) = 1/2

We want to solve for the present disintegration rate, which is given by λN(t).

First, let's solve for N/N0:

1/2 = (1/2)^(t/t1/2)
1/2 = e^(-λt)
ln(1/2) = -λt
λ = ln(2)/t1/2 = ln(2)/5730 years
t = 2.10×10^13 years = 2.10×10^13 years / (365.25 days/year) / (24 hours/day) / (60 minutes/hour) = 3.98×10^20 minutes

Now we can solve for N/N0:

N(t)/N0 = e^(-λt) = e^(-(ln2/5730 years)*(3.98×10^20 minutes))
N(t)/N0 = 0.0123

So the present disintegration rate is:

λN(t) = (ln(2)/5730 years) * 15.3 disintegrations/min/g * 0.0123
λN(t) = 0.000124 disintegrations/min/g

Rounding to one significant figure, we get:

λN(t) = 0.0001 disintegrations/min/g ≈ 0.0001 ≈ 1.2×10^-4 ≈ 0.00012 ≈ 0.00013

The closest answer is B) 11.9.

Assuming That the Original Activity of Carbon - 14 Was