Q. 5.80 - An Introduction to General, Organic, and Biological Chemistry

Question

Write the balanced nuclear equation for each of the following radioactive emissions: $(5.2)$

a. an alpha particle from $Gd-148$

b. a beta particle from $Sr-90$

c. a positron from $\mathrm{Al}-25$

Step-by-Step Solution

Answer

Option a

      As a result, the whole nuclear equation is provided below.

${}_{64}{}^{148}\mathrm{Gd} \to {}_{62}{}^{144}\mathrm{Sm}+{}_2{}^4\mathrm{He}$

Option b

     As a result, the whole nuclear equation is shown below.

${}_{38}{}^{90}\mathrm{Sr}\to {}_{39}{}^{90}\mathrm{Y}+{}_{-1}{}^0\mathrm{e}$

Option c

     As a result, the whole nuclear equation is shown below.

${}_{13}{}^{25}\mathrm{AI}\to {}_{12}{}^{25}\mathrm{Mg}+{}_{-1}{}^0\mathrm{e}$

Part (a) Step 1: Given Information

The mass number of the radioactive nucleus on the left-hand side of a balanced chemical equation is always equal to the sum of the mass numbers of the new nucleus and the released particle on the right-hand side.

Similarly, on the left-hand side of the equation, the atomic number of the radioactive nucleus is equal to the sum of the atomic numbers of the new nucleus and the released particle on the right-hand side.

The alpha particle is a helium nucleus-like nuclear particle.

It is denoted by the symbol ${}_2{}^4\mathrm{He}$.

The mass number is represented by the superscript $\left(4\right)$.

The atomic number is represented by the subscript $\left(2\right)$.

We must determine the balanced nuclear equation for alpha particle emission from radioactive $Gd-148$.

The symbol for gadolinium is $Gd$. It has an atomic number of $64$ and a mass of $148.$ As a result, this radioisotope is written as ${}_{64}{}^{148}Gd$.

Part (a) Step 2: Explanation

The nuclear equation for the alpha particle emission from radioactive $Gd$ is shown below.

 ${}_{64}{}^{148}Gd \to \_\_ +{}_2{}^{4}He$

This equation represents an unfinished chemical reaction.

We must locate the new radioactive nucleus in an empty space.

The mass number of $Gd$ is equal to the sum of the mass numbers of the alpha particle and the new nucleus in the equation above.

Let's say the new nucleus has a mass number of $x$.

Hence,

$148=x+4$

Therefore,

$$\begin{gathered} x=148-4 \\ =144 \end{gathered}$$

As a result, the new nucleus' mass number is $144.$

Part (a) Step 3: Conclusion

$\mathrm{Gd}\text{'}\mathrm{s}$ atomic number is equal to the sum of the new nucleus' and the alpha particle's atomic numbers.

Let's say $y$ is the new nucleus' atomic number.

Therefore,

$64=y+2$

Thus, 

$y=64-2$

Therefore, 

$y=62$

As a result, the new nucleus' atomic number is $62.$

Samarium $Sm$ has the atomic number $62$, according to the periodic table.

As a result, the new nucleus's symbol is ${}_{\mathit{62}}{}^{\mathit{144}}Sm$.

As a result, the whole nuclear equation is provided below.

${}_{64}{}^{148}Gd\to {}_{62}{}^{144}Sm+{}_2{}^{4}He$

Part (b) Step 1: Given Information

(b) The beta particle is an electron-like nuclear particle. When a neutron transforms into a proton and an electron, it forms in the nucleus.

${}_{-1}{}^0\mathrm{He}$ is the symbol for it.

The balanced nuclear equation for the emission of beta from radioactive $Sr-90$ must be found.

The symbol for strontium is $Sr$. It has an atomic number of $38$ and a mass of $90$.

As a result, this radioisotope is written as ${}_{38}{}^{90}\mathrm{Sr}$.

The nuclear equation for the emission of beta from radioactive $Sr$ can be found below.

 ${}_{38}{}^{90}\mathrm{Sr}\to \mathrm{\_\_}+{}_{-1}{}^0\mathrm{e}$

This equation represents an unfinished chemical reaction.

We must locate the new radioactive nucleus in an empty space.

The mass number of $Sr$ is equal to the sum of the mass numbers of the beta particle and the new nucleus in the foregoing equation.

Let's say the new nucleus has a mass number of $x$.

Hence,

$90=x+0$

Therefore,  

$=90$

As a result, the new nucleus' mass number is $90$.

Part (b) Step 2: Explanation

The sum of the atomic numbers of the new nucleus and the beta particle equals the atomic number of $Sr$.

Let's say  $y$ is the new nucleus' atomic number.

Therefore, 

$38=y+(-1)$

Thus, 

$y=38+1$

Therefore,$y=39$

As a result, the new nucleus's atomic number is 39.

The atomic number of yttrium $Y$ is $39$ on the periodic table. As a result, the new nucleus has the symbol ${}_{39}{}^{90}Y$.

As a result, the whole nuclear equation is shown below.

${}_{38}{}^{90}Sr\to {}_{39}{}^{90}Y+{}_{-1}{}^{0}e$

Part (c) Step 1: Given Information

When a proton is converted into a neutron and a positron, a particle with no mass and a positive charge is generated.

It is denoted by the symbol ${}_{-1}{}^0\mathrm{e}$.

The balanced nuclear equation for positron emission from radioactive $Al=25$ must be found.

The symbol for aluminium is $Al$. It has the atomic number $13$ and the mass number $25$. As a result, this radioisotope is designated as ${}_{13}{}^{25}AI$

The nuclear equation for the positron emission from radioactive $Al$ can be found below.

${}_{13}{}^{25}\mathrm{AI}\to \mathrm{\_\_}+{}_{-1}{}^0\mathrm{e}$

This equation represents an unfinished chemical reaction.

Part (c) Step 2: Explanation

We must locate the new radioactive nucleus in an empty space.

The mass number of $Al$ is equal to the sum of the mass numbers of the positron particle and the new nucleus in the previous equation.

Let's say the new nucleus has a mass number of $x$.

Hence,

$25=x+0$

Therefore,

x=25

As a result, the new nucleus' mass number is $25$.

The sum of the atomic numbers of the new nucleus and the proton particle equals the atomic number of $Al$.

Let's say $y$ is the new nucleus' atomic number.

Therefore,

$13=y+1$

Thus,

$y=13-1$

Therefore,

$y=12$

As a result, the new nucleus' atomic number is $12.$

Magnesium $\left(Mg\right)$ has the atomic number $12$ according to the periodic table. As a result, the new nucleus's symbol is ${}_{12}{}^{25}Mg$.

As a result, the whole nuclear equation is shown below.

${}_{13}{}^{25}\mathrm{AI}\to {}_{12}{}^{25}\mathrm{Mg}+{}_{-1}{}^0\mathrm{e}$