91Ó°ÊÓ

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

Age of the rock, t = 260 million years

Mass of ${}^{238}U$in the rock, $m_U=3.70\mathrm{mg}$

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 change in the number of atoms of uranium atoms is also the production of lead atoms through the decay of uranium.

Formulae:

The disintegration constant, $\mathrm{λ}=\frac{\ln 2}{T_{1/2}}…........\left(1\right)$

where, $T_{1/2}$is the half-life of the substance.

The undecayed sample remaining after a given time, $N=N_0{e}^{-\mathrm{λ}t}............\left(2\right)$ The number of undecayed atoms in a given mass of a substance,

$N=\frac{m}{A}{N}_A...........\left(3\right)where{N}_A=6.022×{10}^{23}\mathrm{atoms}/\mathrm{mol}$

Where, A is the molar mass of the substance.

Now, the original amount of present in the rock can be given using equation (3) in equation (2) as follows:

$$\begin{gathered} m_{U0}=m_U{e}^{\mathrm{λ}t}(As,N\mathrm{Î\pm }m) \\ =m_U{e}^{\left(\frac{\ln 2}{T_{1/2}}\right)t}(fromequation\left(1\right)) \\ =\left(3.70mg\right)e^{\left(\frac{\ln 2}{4.47×{10}^{9}y}\right)\left(260×{10}^{6}y\right)} \\ =3.85mg \end{gathered}$$

Thus, the amount of the lead produced can be given as:

$$\begin{gathered} m\text{'}=\left(m_{0U}-m_U\right)\left(\frac{m_{206Pb}}{m_{238U}}\right) \\ =\left(3.85mg-3.70mg\right)\left(\frac{206u}{238u}\right) \\ =0.13mg \end{gathered}$$

Hence, the mass of lead in the rock is 0.13 mg.