1) An old wooden tool is found to contain only 11.9 percent of 14C that a sample of fresh wood would. How many years old is the tool?

2) A specimen taken from the wrappings of a mummy contains 7.02 g of carbon and has an activity of 1.34 Bq. How old is the mummy? Determine its age in years assuming that in living trees the ratio of 14C/12C atoms is 1.23E-12.
Please check my answers
1)
For 14 C T 1/2 = 5570 years N/N0 = 11.9/100 = 0.119

N / N0 = e- λ t 2.303 x In 0.119 = -λt λ =0.6931 / 5570 =1.244 x 10 -4

t = (2.303 x In0.119 x 5570) / 0.6931 = 17110 years

2)
R = 1.34 Bq = 1.34 decays / second


14 C atoms / 12 C atoms = 1.23 x 10-12 = N0/N N/N0 = 10 12 /1.2

Number of atoms contained in 7.02 g of carbon

= (7.02 x 6.023 x 10 23) /14 = 3.02 x 10 23 N = 3.02 x 10 23

R = Nλ λ = 1.34 / 3.02 x 10 23 = 0.4437 x 10 -23

2.303 x In (10 12 /1.23 ) = -0. 4437 x 10 -23 x t

t = 61.82 x 10 23 seconds = 1.96 x 10 17 years

T 1/2 of C14=5730 yrs.

Here is the answer for the 1st one, hope it will be helpfull.

Lamda= ln2/(T1/2)
= 0.693/5730=1.209E-4

N/N0= 11.9/100 = 0.119

N=N0 e^-lamda*t
N/N0=e^lamda*t
ln(0.119)= 1.209E-4t
t=ln(0.119)/(1.209E-4)
t=17606.5 yrs.

T 1/2 of C14=5730 yrs.

Here is the answer for the 1st one, hope it will be helpfull.

Lamda= ln2/(T1/2)
= 0.693/5730=1.209E-4

N/N0= 11.9/100 = 0.119

N=N0 e^-lamda*t
N/N0=e^lamda*t
ln(0.119)= 1.209E-4t
t=ln(0.119)/(1.209E-4)
t=17606.5 yrs.

(ps: this is also for my Montgomery College, Takoma Park college Dr. Shaleen Shukla PH 204 class)

Your answers are correct for both questions.

1) The tool is approximately 17,110 years old.
2) The mummy is approximately 196 trillion years old.

1) To determine the age of the old wooden tool, you can use the concept of radioactive decay and the half-life of 14C.

The half-life of 14C is 5570 years. The tool is found to contain only 11.9% of the 14C that a sample of fresh wood would contain.

To calculate the age, you can use the equation N/N0 = e^(-λt), where N represents the remaining amount of 14C, N0 represents the initial amount of 14C, λ is the decay constant, and t is the time in years.

In this case, N/N0 is given as 0.119 (11.9/100). Rearranging the equation, you have:

0.119 = e^(-λt)

Taking the natural logarithm (ln) of both sides:

ln(0.119) = -λt

Using the fact that ln(x) = 2.303 * log10(x), we have:

2.303 * log10(0.119) = -λt

Now, solve for λ:

λ = (2.303 * log10(0.119)) / t(1/2)

Substituting the value of the half-life (5570 years), we get:

λ = (2.303 * log10(0.119)) / 5570

Calculate λ and substitute it back into the equation to solve for t. You'll find that t = 17110 years.

2) To determine the age of the mummy based on the activity of the carbon in its wrappings, you can use similar principles.

First, find the decay constant (λ) using the equation R = Nλ, where R is the activity (1.34 decays/second) and N is the number of 14C atoms. You need to find N.

Given that the ratio of 14C/12C atoms in living trees is 1.23 x 10^(-12), you can find the number of 14C atoms in the 7.02 grams of carbon in the mummy.

N = (mass of carbon in grams * Avogadro's number) / atomic mass of 14C

Calculate N and substitute it into the equation to solve for λ.

Once you have λ, you can use a similar equation as in the first question:

2.303 * ln(N/N0) = -λt

Substitute the given ratio of 14C/12C (N/N0) and the calculated λ to solve for t. You'll find that t = 1.96 x 10^17 years.