Question

The Destruction of Jericho The Bible describes the Exodus as a period of 40 years that began with plagues in Egypt and ended with the destruction of Jericho. Archeologists seeking to establish the exact dates of these events have proposed that the plagues coincided with a huge eruption of the volcano Thera in the Aegean Sea Radiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC +/- 13 years for the fall of the city. What is the relative amount of ¹⁴C in the old grain vs. the new grain in 2007 AD

(A_t/A_o)_{2007 AD =?}

Answer 1

Exponential growth and decay are rates; i.e they represent the change in some quantity through time.

Exponential growth is any increase in a quantity (N)

Exponential decay is any decrease in a quantity N

through time according to the equations:

N(t) = N₀e^kt (exponential growth)

or

N(t) = N₀e^-kt(exponential decay)

where:

·         N₀is the initial quantity

·         t is time

·         N(t) is the quantity after time t

·         k is a constant (analogous to the decay constant) and

·         e^x is the exponential function (e is the base of the natural logarithm)

Rearrange the equation N_t = N₀ x e^-kt

to get N/N₀ = e^(-kt).

Substitute N and N0 with ¹⁴C_​2007AD and ¹⁴C_1315BC

So we have ¹⁴C_​2007AD / ¹⁴C_{​1315 BC} = e^(-kt)

To find the ratio of C-14 to C-14 in between 1315 BC and 2007 AD years (3322 years)

t = 3322 years

Next,

we need to find the k (rate of decay):

the formula for k

k = ln2 / half-life.

The half life of carbon is 5730 years.

So k = ln2/5730 yr = 1.2097 x 10^-4 yr^-1

Now, substitute the values the formula. we will get

¹⁴C_​2007AD/¹⁴C_{​1315 BC} = e^{-(1.2097 x 10-4)(3322)}

¹⁴C_​2007AD/¹⁴C_{​1315 BC} = 0.6690728

So the ratio of ¹⁴C in comparison of 2007 AD to that of 1315 BC = 0.6690728