Question

In Problems 23–27, assume that the rate of decay of a radioactive substance is proportional to the amount of the substance present. The half-life of a radioactive substance is the time it takes for one-half of the substance to disintegrate.

To see how sensitive the technique of carbon dating of Problem 25 is

(a) Redo Problem 25 assuming the half-life of carbon-14 is 5550 yr.

(b) Redo Problem 25 assuming 3% of the original mass remains.

Short answer

(a) The estimated age of the skull is 31323 years.

(b) The estimated age of the skull is 28330 years.

Step-by-step solution

Given data

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present.

Analyzing the given statement

(a)

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present.

Let the present amount of the radioactive substance be N.

Therefore,

$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{\propto }\mathit{N}$

Given that there areonly 2% of the original amount of carbon-14 remains in the burnt wood of the campfire. We have to estimate theage of the skull if the half-life of carbon-14 is about 5550 years.

Determining the formula with the help of the given proportionality relation, to solve the question

Given,

$\frac{dN}{dt}\propto N$

$\frac{dN}{dt}=-\lambda N$where, $\lambda$is the constant of proportionality.

$$\begin{gathered} \frac{dN}{N}=-\lambda \\ \int \frac{dN}{N}=-\lambda \int dt \\ lnN=-\lambda t+ln{N}_0 \end{gathered}$$

Where, $ln{N}_0$is an arbitrary constant.

$$\begin{gathered} lnN-ln{N}_0=-\lambda t \\ ln\left(\frac{N}{N_0}\right)=-\lambda t \\ \frac{N}{N_0}=e^{-\lambda t} \\ \frac{N}{N_0}=e^{-\lambda t} \end{gathered}$$ …… (1)

One will use this formula to solve the question.

To determine the value of λ

The half-life of carbon-14 is given as 5550 years. The formula for finding the half-life is,

${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{l}\mathbf{n}\mathbf{2}}{\mathbf{\lambda }}$.

Here,$t_{\frac{\mathbf{1}}{\mathbf{2}}}=5550years$

Therefore,$5550=\frac{ln2}{\lambda }$

$\lambda =\frac{ln2}{5550}$ …… (2)

One will use this value of $\lambda$in step4 to find the estimated age of the skull.

To find the estimated age of the skull

Let theoriginal amount of carbon-14be N₀ and let the amount of remaining carbon-14 in the burnt wood of the campfire be N, which is given as 2% of the original amount, i.e., N=0.02N₀

Using the equation (1),

$$\begin{gathered} \left(0.02\right)N_0=N_0{e}^{-\lambda t} \\ 0.02=e^{-\lambda t} \\ {e}^{\lambda t}=\frac{100}{2} \\ {e}^{\lambda t}=50 \\ \lambda t=ln50 \\ t=\frac{ln50}{\lambda } \end{gathered}$$

Now, using the value of role="math" localid="1664199443715" $\mathit{\lambda }\mathbf{}$from equation (2),

$$\begin{gathered} t=\frac{ln50}{ln2}·\left(5550\right) \\ t=31323years \end{gathered}$$

Hence, the estimated age of the skull is31323ears.

Analyzing the given statement

(b)

Given that the rate of decay of a radioactive substance is directly proportional to the amount of the substance present.

Let the present amount of the radioactive substance be N.

Thus,$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{\propto }\mathit{N}$

Given that there areonly 3% of the original amount of carbon-14 remains in the burnt wood of the campfire. We have to estimate theage of the skull if the half-life of carbon-14 is about 5600 years.

Determining the formula with the help of the given proportionality relation, to solve the question

Given,

$\frac{dN}{dt}\propto N$

$\frac{dN}{dt}=-\lambda N$where, $\lambda$is the constant of proportionality.

$$\begin{gathered} \frac{dN}{N}=-\lambda \\ \int \frac{dN}{N}=-\lambda \int dt \\ lnN=-\lambda t+ln{N}_0 \end{gathered}$$

where, In N₀ is an arbitrary constant.

$$\begin{gathered} lnN-ln{N}_0=-\lambda t \\ ln\left(\frac{N}{N_0}\right)=-\lambda t \\ \frac{N}{N_0}=e^{-\lambda t} \\ N=N_0{e}^{-\lambda t} \end{gathered}$$ …… (2)

One will use this formula to solve the question.

 Step 8: To determine the value of λ

The half-life of carbon-14 is given as 5600 years. The formula for finding the half-life is,

${\mathit{t}}_{\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\frac{\mathbf{l}\mathbf{n}\mathbf{2}}{\mathbf{\lambda }}$

Here,$t_{\frac{\mathbf{1}}{\mathbf{2}}}=5600years$

Accordingly,$5600=\frac{ln2}{\lambda }$

$\lambda =\frac{ln2}{5600}$ …… (3)

One will use this value of in step4 to find the estimated age of the skull.

To find the estimated age of the skull

Let theoriginal amount of carbon-14be N₀ and let the amount of remaining carbon-14 in the burnt wood of the campfire be N, which is given as 3% of the original amount, i.e.,$\mathrm{N}=0.03 {\mathrm{N}}_0$

Using the equation (2),

$$\begin{gathered} \left(0.03\right)N_0=N_0{e}^{-\lambda t} \\ 0.03=e^{-\lambda t} \\ {e}^{\lambda t}=\frac{100}{3} \\ \lambda t=ln\left(\frac{100}{3}\right) \\ t=\frac{1}{\lambda }ln\left(\frac{100}{3}\right) \end{gathered}$$

Now, using the value of role="math" localid="1664199348494" $\mathit{\lambda }$from equation (2),

$$\begin{gathered} t=\frac{ln\left(\frac{100}{3}\right)}{ln2}·\left(5600\right) \\ t=28330years \end{gathered}$$

So, the estimated age of the skull is 28330years.