Chapter 20, Problem 116IP

Interpretation Introduction

Interpretation: It is given that, radon is present as $1.0\text{g}$ per $7.0\text{metric}\text{ton}$ of a pitch ball. The number of radon atoms that can be isolated from given mass of pitch-blende is to be calculated. The number of radon atoms remain in $2025$ is to be calculated if $15\text{mg}$ radon is manufactured in $1925$.

Concept introduction: Amount of radon left is calculated using the formula,

$A_{\text{E}}=A_0{0.5}^{t/t_{1/2}}$

To determine: The number of radon atoms that can be isolated from $1.75\times {10}^8\text{g}$ pitch-blende; the number of radon atoms remain in $2025$, if $15\text{mg}$ radon is manufactured in $1925$.

Answer to Problem 116IP

Answer

Number of radon $\left(\text{Ra}\right)$ atoms that can be isolated from $1.75\times {10}^8\text{g}$ pitch-blende is $\underset{\_}{6.66\times 10^{22}atoms}$. Number of radon $\left(\text{Ra}\right)$ atoms left in $2025$ is $\underset{\_}{3.82\times 10^{19}Raatoms}$.

Explanation of Solution

Mass of radon $\left(\text{Ra}\right)$ is $\underset{\_}{25g}$.

Given

Mass of pitch-blende is $1.75\times {10}^8\text{g}$.

The conversion of gram $\left(\text{g}\right)$ into kilogram $\left(\text{kg}\right)$ is done as,

$\text{1}\text{g}={\text{10}}^{-\text{3}}\text{kg}$

Hence,

The conversion of $1.75\times {10}^8\text{g}$ into kilogram is,

$\begin{array}{c}\text{1}\text{.75}\times {\text{10}}^{\text{8}}\text{g}=\left(\text{1}\text{.75}\times {\text{10}}^{\text{8}}\times {\text{10}}^{-\text{3}}\right)\text{kg}\\ =\text{1}\text{.75}\times {\text{10}}^{\text{5}}\text{kg}\end{array}$

Since, radon is present as $1.0\text{g}$ per $7.0\text{metric}\text{ton}$ of a pitch ball, mass of radon is calculated as,

$\text{Mass}\text{of}\text{Ra}=\text{Mass}\text{of}\text{pitch-blende}\times \frac{\text{1}\text{.0}\text{g}\text{Ra}}{\text{7}\text{.0}\text{metric}\text{ton}}\times \left(\frac{\text{1}\text{metric}\text{ton}}{\text{1000}\text{kg}}\right)$

Substitute the value of mass of pitch-blende in the above equation.

$\begin{array}{c}\text{Mass}\text{of}\text{Ra}=\text{Mass}\text{of}\text{pitch-blende}\times \frac{\text{1}\text{.0}\text{g}\text{Ra}}{\text{7}\text{.0}\text{metric}\text{ton}}\times \left(\frac{\text{1}\text{metric}\text{ton}}{\text{1000}\text{kg}}\right)\\ =\text{1}\text{.75}\times {\text{10}}^{\text{5}}\text{kg}\times \frac{\text{1}\text{.0}\text{g}\text{Ra}}{\text{7}\text{.0}\text{metric}\text{ton}}\times \left(\frac{\text{1}\text{metric}\text{ton}}{\text{1000}\text{kg}}\right)\\ =\underset{\_}{25g}\end{array}$

Number of radon $\left(\text{Ra}\right)$ atoms that can be isolated from $1.75\times {10}^8\text{g}$ pitch-blende is $\underset{\_}{6.66\times 10^{22}atoms}$.

Mass of radon is $25\text{g}$.

Atomic mass of radon is $226\text{g}$.

Formula

The number of radon atoms is calculated as,

$\text{Atoms}\text{of}\text{Ra}=\text{Mass}\text{of}\text{Ra}\times \frac{1\text{mol}\text{Ra}}{\text{Atomic}\text{mass}\text{of}\text{Ra}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}$

Substitute the values of mass and atomic mass of radon in the above equation.

$\begin{array}{c}\text{Atoms}\text{of}\text{Ra}=\text{Mass}\text{of}\text{Ra}\times \frac{1\text{mol}\text{Ra}}{\text{Atomic}\text{mass}\text{of}\text{Ra}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}\\ =25\text{g}\times \frac{1\text{mol}\text{Ra}}{\text{226}\text{.0}\text{g}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}\\ =\underset{\_}{6.66\times 10^{22}atoms}\end{array}$

Number of radon $\left(\text{Ra}\right)$ atoms in $15\text{mg}$ is $\underset{\_}{3.99\times 10^{19}atoms}$.

Given

Mass of radon atoms is $15\text{mg}$.

Half life of radon is $1.60\times {10}^3\text{years}$.

Atomic mass of radon is $226\text{g}$.

The conversion of milligram $\left(\text{mg}\right)$ into gram $\left(\text{g}\right)$ is done as,

$\text{1}\text{mg}={\text{10}}^{-\text{3}}\text{g}$

Hence,

The conversion of $15\text{mg}$ into gram is,

$\begin{array}{c}\text{15}\text{mg}=\left(\text{15}\times {\text{10}}^{-\text{3}}\right)\text{g}\\ =\text{15}\times {\text{10}}^{-\text{3}}\text{g}\end{array}$

Formula

The number of radon atoms is calculated as,

$\text{Atoms}\text{of}\text{Ra}=\text{Mass}\text{of}\text{Ra}\times \frac{1\text{mol}\text{Ra}}{\text{Atomic}\text{mass}\text{of}\text{Ra}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}$

Substitute the values of mass and atomic mass of radon in the above equation.

$\begin{array}{c}\text{Atoms}\text{of}\text{Ra}=\text{Mass}\text{of}\text{Ra}\times \frac{1\text{mol}\text{Ra}}{\text{Atomic}\text{mass}\text{of}\text{Ra}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}\\ =\text{15}\times {\text{10}}^{-\text{3}}\text{g}\times \frac{1\text{mol}\text{Ra}}{\text{226}\text{.0}\text{g}}\times \frac{6.022\times {10}^{23}\text{atoms}}{1\text{mol}\text{Ra}}\\ =\underset{\_}{3.99\times 10^{19}atoms}\end{array}$

Number of radon $\left(\text{Ra}\right)$ atoms left in $2025$ is $\underset{\_}{3.82\times 10^{19}Raatoms}$.

Half life of radon is $1.60\times {10}^3\text{years}$.

Number of radon $\left(\text{Ra}\right)$ atoms in $15\text{mg}$ is $\text{3}\text{.99}\times {\text{10}}^{\text{19}}\text{atoms}$.

Total time from $1925$ to $2025$ is,

$2025-1925=100\text{years}$

Formula

Amount of radon left is calculated using the formula,

$A_{\text{E}}=A_0{0.5}^{t/t_{1/2}}$

Where,

• $A_{\text{E}}$ is the amount of radon left.
• $A_{\text{0}}$ is the original amount of radon.
• $t_{1/2}$ is the half-life.
• $t$ is the total time.

Substitute the values of $t_{1/2},t$ and $A_{\text{0}}$ in the above equation.

$\begin{array}{c}{A}_{\text{E}}=A_0{0.5}^{t/t_{1/2}}\\ =3.99\times {10}^{19}\text{Ra}\text{atoms}\times {\left(0.5\right)}^{\frac{100\text{years}}{1600\text{years}}}\\ =\underset{\_}{3.82\times 10^{19}Raatoms}\end{array}$

Conclusion

Conclusion

The calculated value of number of radon $\left(\text{Ra}\right)$ atoms left in $2025$ is $\underset{\_}{3.82\times 10^{19}Raatoms}$. Number of radon $\left(\text{Ra}\right)$ atoms in $25\text{g}$ is $\underset{\_}{6.66\times 10^{22}atoms}$