In nature, potassium has three isotopes, $^ { 3 9 } \mathrm { K}$ , $^ { 4 0 } \mathrm { K}$ and $^ { 4 1 } \mathrm { { K } }$, with abundances of $93.2581\%$, $0.0117\%$ and $6.7302\%$ respectively. $^ { 3 9 } \mathrm { { K } }$ and $^ { 41 } \mathrm { { K } }$ are stable nuclides; while $^ { 40 } \mathrm { { K } }$ is radioactive, and its decay may produce $^ { 40 } \mathrm { C a }$ (partial half-life $t _ { 1 } = 1 . 3 9 7 \times 1 0 ^ { 9 } \ \rm{y} ) $ or $^ { 40 } \mathrm { A r ( }$ partial half-life $t _ { 2 } = 1 . 1 9 3 \times 1 0 ^ { 10 } \ \mathrm { y } $). Argon has three isotopes, $^ { 3 6 } \mathrm { A r }$, ${ } ^ { 3 8 } \mathrm { A r }$ and $^ { 40 } \mathrm { A r }$, with abundances of $0.334\%$, $0.063\%$ and $99.604\%$ respectively. $^ { 3 6 } \mathrm { A r }$, $^ { 3 8 } \mathrm { A r }$ and $^ { 40 } \mathrm { A r }$ are all stable nuclides. Granite is a rock formed by the cooling and solidification of magma from volcanic eruptions. When granite forms, it seals a very small amount of atmosphere inside the rock. These gases cannot diffuse out of the rock unless they are completely released by heating and melting at high temperatures. To determine the age of a piece of granite, a $1.000 \ \mathrm { g }$ granite sample was thoroughly heated and melted to completely release the argon gas inside; the released gas was mixed with $n _ { 1 } = 8 . 3 0 \times 1 0 ^ { - 1 0 }\ \rm{mol}$ of pure ${ ^{38} \rm{ Ar} }$, and the mass spectrometer measured $n(^{38}\mathrm{Ar}):n(^{36}\mathrm{Ar})=371.8$ and $n(^{40}\mathrm{Ar}):n(^{38}\mathrm{Ar})=1.400$ in the mixed gas. Another sample was taken, and atomic absorption spectroscopy was used to measure the potassium content as $6.065\%$. Assume that the isotopic abundances of potassium and argon in nature do not change with age, and the following options are closest to the age of the granite sample
在自然界中，钾有 $^ { 3 9 } \mathrm { K}$ ，$^ { 4 0 } \mathrm { K}$ 和 $^ { 4 1 } \mathrm { { K } }$ 三种同位素，其丰度分别为 $9 3 . 2 5 8 1 \%$ 、$0.0117\%$和$ 6 . 7 3 0 2 \%$ 。$^ { 3 9 } \mathrm { { K } }$ 和 $^ { 4 1 } \mathrm { { K } }$ 为稳定核素；而 $^ { 4 0 } \mathrm { { K } }$ 有放射性，其衰变可能生成 $^ { 4 0 } \mathrm { C a }$ （分支半衰期 $t _ { 1 } = 1 . 3 9 7 \times  1 0 ^ { 9 } \ \rm{y} ) $ 或 $^ { 4 0 } \mathrm { A r ( }$ 分支半衰期 $t _ { 2 } = 1 . 1 9 3 \times 1 0 ^ { 1 0 } \ \mathrm { y } $） 。氩有 $^ { 3 6 } \mathrm { A r }$ ，${ } ^ { 3 8 } \mathrm { A r }$ 和 $^ { 4 0 } \mathrm { A r }$ 三种同位素，其丰度分别为$0.334\%$、 $0 . 0 6 3 \%$ 和 $9 9 . 6 0 4 \%$。$^ { 3 6 } \mathrm { A r }$ ， $^ { 3 8 } \mathrm { A r }$ 和 $^ { 4 0 } \mathrm { A r }$ 均为稳定核素。花岗岩是一种由火山爆发的岩浆冷却凝固形成的岩石。花岗岩成型时，会将极少量的大气封存在岩石内部。这些气体无法扩散出岩石之外，只有经过高温加热熔融才能完全放出。为测定一块花岗岩的年龄，取花岗岩样品 $1 . 0 0 0 \ \mathrm { g }$ 充分加热熔融，使其中的氩气完全放出；将放出的气体与 $n _ { 1 } = 8 . 3 0 \times 1 0 ^ { - 1 0 }\ \rm{mol}$的纯 ${ ^{3 8} \rm{ Ar} }$ 混合，用质谱仪测得混合气体中$n(^{38}\mathrm{Ar}):n(^{36}\mathrm{Ar})=371.8$；$n(^{40}\mathrm{Ar}):n(^{38}\mathrm{Ar})=1.400$。另取一份样品，用原子吸收光谱法测得其中含钾 $6 . 0 6 5 \%$ 。设自然界中钾和氩的各同位素丰度不随年代而变化，下列选项中，和该花岗岩样品的年龄最接近的为
$1\times10^{3}\ \rm{y}$
$5\times10^{3}\ \rm{y}$
$1\times10^{4}\ \rm{y}$
$5\times10^{4}\ \rm{y}$
$1\times10^{5}\ \rm{y}$
$5\times10^{5}\ \rm{y}$
$1\times10^{6}\ \rm{y}$
$5\times10^{6}\ \rm{y}$
$1\times10^{7}\ \rm{y}$
$5\times10^{7}\ \rm{y}$
$1\times10^{8}\ \rm{y}$
$5\times10^{8}\ \rm{y}$
$1\times10^{9}\ \rm{y}$
$5\times10^{9}\ \rm{y}$
$1\times10^{10}\ \rm{y}$
$1\times10^{3}\ \rm{y}$
$5\times10^{3}\ \rm{y}$
$1\times10^{4}\ \rm{y}$
$5\times10^{4}\ \rm{y}$
$1\times10^{5}\ \rm{y}$
$5\times10^{5}\ \rm{y}$
$1\times10^{6}\ \rm{y}$
$5\times10^{6}\ \rm{y}$
$1\times10^{7}\ \rm{y}$
$5\times10^{7}\ \rm{y}$
$1\times10^{8}\ \rm{y}$
$5\times10^{8}\ \rm{y}$
$1\times10^{9}\ \rm{y}$
$5\times10^{9}\ \rm{y}$
$1\times10^{10}\ \rm{y}$
$n(^{36}\mathrm{Ar})$ is a stable nuclide that can react to sequester the total amount of $\rm{Ar}$ in the gas. Suppose the total amount of $\rm{Ar}$ in the sequestered gas is $x\ \rm{mol}$, then we have the equation:

$\frac{0.00063x+8.30\times 10^{-10}}{0.00334x}=371.8$

Solving for $x=6.69\times 10^{-10}$, the total amount of $\rm{Ar}$ is $\rm 6.69\times 10^{-10} ~mol$

<Checkpoint>[The total amount of $\rm{Ar}$ is $\rm 6.69\times 10^{-10}~mol$](1)</Checkpoint>


Since $n(^{38}\mathrm{Ar})=8.30\times 10^{-10}\ \rm{mol}$, we can calculate $n(^{40}\mathrm{Ar})=1.4n(^{38}\mathrm{Ar})=1.16\times 10^{-9}\ \rm{mol}$

From the isotopic abundance, $n(^{40}\mathrm{Ar})_1$ from $\rm{Ar}$ itself is $0.99604x=6.67\times 10^{-10}\ \rm{mol}$

<Checkpoint>[$n(^{40}\mathrm{Ar})_1$ from $\rm{Ar}$ itself is $0.99604x=6.67\times 10^{-10}\ \rm{mol}$](1)</Checkpoint>

Therefore, $n(^{40}\mathrm{Ar})_2$ from $^{40}\mathrm{K}$ is $n(^{40}\mathrm{Ar})-n(^{40}\mathrm{Ar})_1=4.93\times 10^{-10}\ \rm{mol}$

<Checkpoint>[$n(^{40}\mathrm{Ar})_2$ from $^{40}\mathrm{K}$ is $n(^{40}\mathrm{Ar})-n(^{40}\mathrm{Ar})_1=4.93\times 10^{-10}\ \rm{mol}$](1)</Checkpoint>

According to atomic absorption spectroscopy, and using $39.10$ to approximate the weighted average atomic mass of potassium, we can calculate the total amount of $\rm{K}$, $n(\rm{K})=\frac{1.000\times0.06065}{39.10}=0.001551\ \rm{mol}$

<Checkpoint>[Total $\rm{K}$, $n(\rm{K})=\frac{1.000\times0.06065}{39.10}=0.001551\ \rm{mol}$](1)</Checkpoint>

Then $n(^{40}\rm{K})=0.000117n(\rm{K})=1.81\times10^{-7}\ \rm{mol}$

<Checkpoint>[$n(^{40}\rm{K})=0.000117n(\rm{K})=1.81\times10^{-7}\ \rm{mol}$](1)</Checkpoint>

The decay reaction is a first-order reaction, and the rate constant $k$ can be obtained from the half-life:

$k_1=\frac{\ln2}{t_1}=4.962\times 10^{-10}\ \rm{y^{-1}}$

<Checkpoint>[$k_1=\frac{\ln2}{t_1}=4.962\times 10^{-10}\ \rm{y^{-1}}$](1)</Checkpoint>

$k_2=\frac{\ln2}{t_2}=5.810\times 10^{-11}\ \rm{y^{-1}}$

<Checkpoint>[$k_2=\frac{\ln2}{t_2}=5.810\times 10^{-11}\ \rm{y^{-1}}$](1)</Checkpoint>

Then the consumption rate of $^{40}\rm{K}$ can be expressed as:

$\frac{-dn(^{40}\mathrm{K})}{dt}=k_1n(^{40}\mathrm{K})+k_2n(^{40}\mathrm{K})=(k_1+k_2)n(^{40}\mathrm{K})$

Integrating, we have $\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$

<Checkpoint>[$\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$](1)</Checkpoint>

Also, the total amount of $^{40}\mathrm{K}$ consumed is $n(^{40}\mathrm{Ar})_2\times\frac{k_1+k_2}{k_2}=4.70\times 10^{-9}\ \rm{mol}$

<Checkpoint>[The total amount of $^{40}\mathrm{K}$ consumed is $n(^{40}\mathrm{Ar})_2\times\frac{k_1+k_2}{k_2}=4.70\times 10^{-9}\ \rm{mol}$](1)</Checkpoint>

Then $\rm n(^{40}\mathrm{K})_{end}=n(^{40}\rm{K})-4.70\times 10^{-9} = 1.76\times 10^{-7} ~mol$

Substituting each value into $\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$, solving for $t=4.74\times 10^7\ \rm{y}$

<Checkpoint>[$t=4.74\times 10^7\ \rm{y}$](1)</Checkpoint>
$n(^{36}\mathrm{Ar})$是一种稳定核素，可以反应封存气体中$\rm{Ar}$的总量。不妨假设封存气体中$\rm{Ar}$的总量为$x\ \rm{mol}$，则有方程：

$\frac{0.00063x+8.30\times 10^{-10}}{0.00334x}=371.8$

解出$x=6.69\times 10^{-10}$,$\rm{Ar}$的总量为$\rm 6.69\times 10^{-10} ~mol$

<Checkpoint>[$\rm{Ar}$的总量为$\rm 6.69\times 10^{-10}~mol$](1)</Checkpoint>


由于$n(^{38}\mathrm{Ar})=8.30\times 10^{-10}\ \rm{mol}$，可以计算$n(^{40}\mathrm{Ar})=1.4n(^{38}\mathrm{Ar})=1.16\times 10^{-9}\ \rm{mol}$

由同位素丰度，来自$\rm{Ar}$本身的$n(^{40}\mathrm{Ar})_1=0.99604x=6.67\times 10^{-10}\ \rm{mol}$

<Checkpoint>[来自$\rm{Ar}$本身的$n(^{40}\mathrm{Ar})_1=0.99604x=6.67\times 10^{-10}\ \rm{mol}$](1)</Checkpoint>

因此，来自$^{40}\mathrm{K}$的$n(^{40}\mathrm{Ar})_2=n(^{40}\mathrm{Ar})-n(^{40}\mathrm{Ar})_1=4.93\times 10^{-10}\ \rm{mol}$

<Checkpoint>[来自$^{40}\mathrm{K}$的$n(^{40}\mathrm{Ar})_2=n(^{40}\mathrm{Ar})-n(^{40}\mathrm{Ar})_1=4.93\times 10^{-10}\ \rm{mol}$](1)</Checkpoint>

根据原子吸收光谱，并用$39.10$近似代替钾的加权平均原子质量，可以计算$\rm{K}$总量$n(\rm{K})=\frac{1.000\times0.06065}{39.10}=0.001551\ \rm{mol}$

<Checkpoint>[$\rm{K}$总量$n(\rm{K})=\frac{1.000\times0.06065}{39.10}=0.001551\ \rm{mol}$](1)</Checkpoint>

则$n(^{40}\rm{K})=0.000117n(\rm{K})=1.81\times10^{-7}\ \rm{mol}$

<Checkpoint>[$n(^{40}\rm{K})=0.000117n(\rm{K})=1.81\times10^{-7}\ \rm{mol}$](1)</Checkpoint>

衰变反应为一级反应，可以根据半衰期求出速率常数$k$：

$k_1=\frac{\ln2}{t_1}=4.962\times 10^{-10}\ \rm{y^{-1}}$

<Checkpoint>[$k_1=\frac{\ln2}{t_1}=4.962\times 10^{-10}\ \rm{y^{-1}}$](1)</Checkpoint>

$k_2=\frac{\ln2}{t_2}=5.810\times 10^{-11}\ \rm{y^{-1}}$

<Checkpoint>[$k_2=\frac{\ln2}{t_2}=5.810\times 10^{-11}\ \rm{y^{-1}}$](1)</Checkpoint>

则$^{40}\rm{K}$的消耗速率可以表示为：

$\frac{-dn(^{40}\mathrm{K})}{dt}=k_1n(^{40}\mathrm{K})+k_2n(^{40}\mathrm{K})=(k_1+k_2)n(^{40}\mathrm{K})$

积分，有$\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$

<Checkpoint>[$\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$](1)</Checkpoint>

又$^{40}\mathrm{K}$的消耗总量为$n(^{40}\mathrm{Ar})_2\times\frac{k_1+k_2}{k_2}=4.70\times 10^{-9}\ \rm{mol}$

<Checkpoint>[$^{40}\mathrm{K}$的消耗总量为$n(^{40}\mathrm{Ar})_2\times\frac{k_1+k_2}{k_2}=4.70\times 10^{-9}\ \rm{mol}$](1)</Checkpoint>

则$\rm n(^{40}\mathrm{K})_{end}=n(^{40}\rm{K})-4.70\times 10^{-9} = 1.76\times 10^{-7} ~mol$

将各数值代入$\ln{\frac{n(^{40}\mathrm{K})}{n(^{40}\mathrm{K})_{end}}}=(k_1+k_2)t$，解出$t=4.74\times 10^7\ \rm{y}$

<Checkpoint>[$t=4.74\times 10^7\ \rm{y}$](1)</Checkpoint>