91Ó°ÊÓ

a) Half-life${}^{238}U$ of$T_{1/2}=4.47×{10}^{9}y$ ,

b) Mass of ${}^{238}U$that the rock contains,$m_u=4.20mg$

c) Mass of ${}^{206}Pb$that the rock contains,$m_{pb}=2.135mg$

d) The rock contains no lead at the formation.

The rock at the time of formation was having only uranium and thus with every decay of one uranium atom, we get one lead atom, thus, the initial atoms present were that of uranium which is the combination of the present amount of combined uranium and lead atoms present in the rock.

The disintegration constant as:

$\mathrm{λ}=\frac{\ln 2}{T_{{\displaystyle \frac{1}{2}}}}$ …… (i)

Here,$T_{\frac{1}{2}}$ is the half-life of the substance.

The undecayed sample remaining after a given time is as follows:

$N=N_0{e}^{-\mathrm{λ}t}$ …… (ii)

The number of undecayed atoms in a given mass of a substance is as follows:

$N=\frac{m}{A}{N}_A$ …â¶Ä¦(¾±¾±¾±)

Here, A is the molar mass of the substance and$N_A=6.022×{10}^{23}\frac{atoms}{mol}$ .

Using the given in equation (iii), the number of${}^{238}U$ atoms present in the rock now can be given as follows:

$$\begin{gathered} N_u=\frac{4.20×{10}^{-3}g}{238{\displaystyle \frac{g}{mol}}}\left(6.022×{10}^{23}\frac{atoms}{mol}\right) \\ =1.06×{10}^{19} \end{gathered}$$

Hence, the number of uranium atoms is$1.06×{10}^{19}$ .

Using the given in equation (iii), the number of${}^{206}Pb$ atoms present in the rock now can be given as follows:

$$\begin{gathered} N_{Pb}=\frac{2.135×{10}^{-3}g}{206{\displaystyle \frac{g}{mol}}}\left(6.022×{10}^{23}\frac{atoms}{mol}\right) \\ =6.24×{10}^{18} \end{gathered}$$

Hence, the number of lead atoms is$6.24×{10}^{18}$ .

If no lead was lost, there was originally one uranium atom for each lead atom formed by decay, in addition to the uranium atoms that did not yet decay. Thus, the original number of uranium atoms was

$N_{U0}=N_U+N_{Pb}$

Substitute the values and solve as:

$$\begin{gathered} N_{U0}=1.06×{10}^{19}+6.24×{10}^{18} \\ =1.684×{10}^{19} \end{gathered}$$

Hence, the number of uranium atoms is$1.684×{10}^{19}$ .

Now, the age of the rock can be calculated by substituting the disintegration constant value of equation (i) in equation (ii) as follows: (considering the uranium decay that is the decay of the uranium atoms)

$$\begin{gathered} N_U=N_{U0}{e}^{-\left(\frac{\ln 2}{T_{1/2}}\right)t} \\ t=\frac{T_{{\displaystyle \frac{1}{2}}}}{\ln 2}\ln \left(\frac{N_U}{N_{U0}}\right) \\ t=-\frac{4.47×{10}^{9}y}{\ln 2}\ln \left(\frac{1.06×{10}^{19}}{1.684×{10}^{19}}\right) \\ t=2.97×{10}^{9}y \end{gathered}$$

Hence, the age of the rock is$2.97×{10}^{9}y$ .