91影视

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

(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,

$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{鈭�}\mathit{N}$

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.

Given,

$\frac{dN}{dt}鈭�N$

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

$$\begin{gathered} \frac{dN}{N}=-位 \\ 鈭�\frac{dN}{N}=-位鈭�dt \\ lnN=-位t+ln{N}_0 \end{gathered}$$

$$\begin{gathered} lnN-ln{N}_0=-位t \\ ln\left(\frac{N}{N_0}\right)=-位t \\ \frac{N}{N_0}=e^{-位t} \\ \frac{N}{N_0}=e^{-位t} \end{gathered}$$ 鈥︹�� (1)

One will use this formula to solve the question.

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

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

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

Therefore,$5550=\frac{ln2}{位}$

$位=\frac{ln2}{5550}$ 鈥︹�� (2)

One will use this value of $位$in step4 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),

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

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

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

Hence, the estimated age of the skull is31323ears.

(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,$\frac{\mathbf{d}\mathbf{N}}{\mathbf{d}\mathbf{t}}\mathbf{鈭�}\mathit{N}$

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.

Given,

$\frac{dN}{dt}鈭�N$

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

$$\begin{gathered} \frac{dN}{N}=-位 \\ 鈭�\frac{dN}{N}=-位鈭�dt \\ lnN=-位t+ln{N}_0 \end{gathered}$$

where, In N₀ is an arbitrary constant.

$$\begin{gathered} lnN-ln{N}_0=-位t \\ ln\left(\frac{N}{N_0}\right)=-位t \\ \frac{N}{N_0}=e^{-位t} \\ N=N_0{e}^{-位t} \end{gathered}$$ 鈥︹�� (2)

One will use this formula to solve the question.

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

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

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

Accordingly,$5600=\frac{ln2}{位}$

$位=\frac{ln2}{5600}$ 鈥︹�� (3)

One will use this value of in step4 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.,$\mathrm{N}=0.03鈥�{\mathrm{N}}_0$

Using the equation (2),

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

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

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

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