Chapter 10, Problem 10.51E

Interpretation Introduction

Interpretation:

The mass in grams of the original sample and mass of the sample after $138$ days are to be calculated.

Concept introduction:

A radioactive decay series is a series of nuclear reaction in which an unstable radioisotope decays to form a stable isotope.

Nuclear reaction deals with the exchange and release of nucleons and changes the identity of the participating elements.

Answer to Problem 10.51E

The mass of the original sample is $0.21\text{g}$. The mass of the sample after $138$ days is $0.\text{208}\text{g}$.

Explanation of Solution

The type of radioactive decay in which an alpha particle is emitted by the nucleus of an atom such that an atom of another element is produced after decay is known as alpha decay. The emission of an alpha particle decreases the atomic number of the parent element by two and atomic mass by four.

The atomic number of $\text{Po-}210$ is $84$. Thus, due to alpha decay the atomic number of daughter nuclei is,

$\begin{array}{rll}& =& 84-2\\ & =& 82\end{array}$

Similarly, the mass umber changes from $210$ to $206$. $\text{Pb}$ is an element whose atomic number is $84$.

Thus, the balanced equation for the decay of $\text{Po-}210$ is as follows:

${}_{84}^{210}\text{Po}{\to }_2^4\text{α}{+}_{82}^{206}\text{Pb}$

It is given that the original sample of $\text{Po-}210$ contains $6.02\times {10}^{20}$ atoms.

The formula to calculate number of atoms is,

$\text{Number}\text{of}\text{atoms}=n\times {\text{N}}_{\text{A}}$

Where,

• $n$ is the number of moles.

• ${\text{N}}_{\text{A}}$ is the Avogadro number.

Substitute the given values in the above formula.

$\begin{array}{rll}\text{Number}\text{of}\text{atoms}& =& n\times {\text{N}}_{\text{A}}\\ 6.02\times {10}^{20}& =& n\times 6.02\times {10}^{23}\\ n& =& \frac{6.02\times {10}^{20}}{6.02\times {10}^{23}}\\ & =& 1\times {10}^{-3}\text{mol}\end{array}$

The formula to calculate number of moles is,

$n=\frac{m}{M_w}$

Where,

• $m$ is the mass of the substance.

• $M_w$ is the molecular weight.

The molecular weight of $\text{Po-}210$ is $210\text{g}/\text{mol}$.

Substitute the value of moles and molar mass in the above formula.

$\begin{array}{rll}n& =& \frac{m}{M_w}\\ 1\times {10}^{-3}\text{mol}& =& \frac{m}{210\text{g}/\text{mol}}\\ m& =& \left(1\times {10}^{-3}\text{mol}\right)\times \left(210\text{g}/\text{mol}\right)\\ & =& 0.21\text{g}\end{array}$

Thus, the mass of the original sample is $0.21\text{g}$.

The given half-life of $\text{Po-}210$ is $138$ days. The half-life period of an element is the amount of time that is required for a radioactive element to get converted to $50\%$ of its initial concentration. Thus, the remaining grams of $\text{Po-}210$ is,

$\begin{array}{l}=\frac{0.21\text{g}}{2}\\ =0.105\text{g}\end{array}$

But $\text{Pb-}206$ is also present, thus; the amount $\text{Pb-}206$ is present in the sample is calculated as follows:

$\begin{array}{rll}\text{Amount}\text{of}\text{Pb-206}& =& 0.105\text{g}\text{Po}\times \frac{\text{1}\text{mol}\text{Po}}{\text{210}\text{g}\text{Po}}\times \frac{\text{1}\text{mol}\text{Pb}}{\text{1}\text{mol}\text{Po}}\times \frac{\text{206}\text{g}\text{Pb}}{\text{1}\text{mol}\text{Pb}}\\ & =& 0.103\text{g}\end{array}$

The total mass of the sample is calculated as follows:

$\begin{array}{rll}\text{Total}\text{mass}& =& \text{Mass}\text{of}\text{Po-210}+\text{Mass}\text{of}\text{Pb-206}\\ & =& 0.105\text{g}+0.103\text{g}\\ & =& 0.\text{208}\text{g}\end{array}$

Thus, the mass of the sample after $138$ days is $0.\text{208}\text{g}$.

Conclusion

The mass of the original sample is $0.21\text{g}$. The mass of the sample after $138$ days is $0.\text{208}\text{g}$.