Chapter 45, Problem 45SP

(a)

To determine

The atomic number and mass number of the nucleus UX₁.

Answer to Problem 45SP

Solution:

$90$, $234$

Explanation of Solution

Given data:

By natural radioactivity, $\text{U}$ emits an alpha particle to form a nucleusUX₁.

The heavy nucleus UX₁ emits a beta particle to form the resultant nucleusUX₂.

Formula used:

The conservation of mass is applied in the reaction $\text{P}\to \text{Q}+\text{R}$ by equating the atomic mass on the left side with the atomic mass on the right side and by equating the atomic number on the left side with the sum of atomic numbers on the right side.

$\begin{array}{l}{Z}_1=Z_2+Z_3\\ A_1=A_2+A_3\end{array}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{P}$, $\text{Q}$, and $\text{R}$, respectively. Here, $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{P}$, $\text{Q}$, and $\text{R}$, respectively.

The reaction, when an alpha particle is emitted from a nucleus, is written as

$\text{P}\to \text{D}+\alpha$

Here, $\alpha$ represents the nucleus of an alpha particle, $\text{P}$ is the parent nucleus with atomic number $Z$ and atomic mass $A$, and $\text{D}$ is the daughter nucleus with atomic number $y$ and mass number x.

Explanation:

Consider the given nucleus $\text{U}$. The symbol $\text{U}$ indicates that this is a uranium isotope. Write the value of the atomic number of uranium from the periodic table. The atomic number of uranium is 92. The complete representation of the nucleus is $\text{U}$.

Compare the nucleus $\text{U}$ with the parent nucleus $\text{P}$. Therefore,

$\begin{array}{c}{Z}_1=92\\ A_1=238\end{array}$

Write the standard expression for the emission of an alpha particle from a nucleus as

$\text{P}\to \text{D}{}_1+\alpha$

Here, $x$ and $y$ indicate the mass number and atomic number of the nucleus UX₁, respectively.

The parent nucleus for the reaction is $\text{U}$ and the daughter nucleus formed is UX₁. Therefore, substitute $\text{U}$ for $\text{P}$, $238$ for $A_1$, $92$ for $N_1$, and ${\text{UX}}_1$ for ${\text{D}}_1$.

$\text{U}\to \text{U}{\text{X}}_1+\alpha$…… (1)

Apply the conservation of mass by equating the atomic masses on both sides of the chemical equation (1):

$\begin{array}{c}238=x+4\\ x=238-4\\ x=234\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of the chemical equation (1):

$\begin{array}{c}92=y+2\\ y=92-2\\ y=90\end{array}$

The value of $x$ is calculated as $234$ and the value of $y$ is calculated as $90$.

Rewrite equation (1) as

$\text{U}\to \text{U}{\text{X}}_1+\alpha$

Substitute $234$ for $x$ and $90$ for $y$

$\text{U}\to \text{U}{\text{X}}_1+\alpha$

Conclusion:

The complete equation, after filling the values of $x$ and $y$, is $\text{U}\to \text{D}{}_1+\alpha$. The daughter nucleus is represented as $\text{U}{\text{X}}_1$, which indicates that the atomic number of the daughter nucleus ${\text{UX}}_1$ is $90$ and the mass number of ${\text{UX}}_1$ is $234$.

(b)

To determine

The atomic number and mass number of the nucleus ${\text{UX}}_2$.

Answer to Problem 45SP

Solution:

$91$, $234$

Explanation of Solution

Given data:

By natural radioactivity, $\text{U}$ emits an alpha particle to form a nucleus ${\text{UX}}_1$.

The heavy nucleus ${\text{UX}}_1$ emits a beta particle to form the resultant nucleus ${\text{UX}}_2$.

Formula used:

The conservation of mass is applied in the reaction $\text{P}\to \text{Q}+\text{R}$ by equating the atomic mass on the left side with the atomic mass on the right side and by equating the atomic number on the left side with the sum of atomic numbers on the right side.

$\begin{array}{l}{Z}_1=Z_2+Z_3\\ A_1=A_2+A_3\end{array}$

Here, $Z_1$, $Z_2$, and $Z_3$ represent the atomic numbers of $\text{P}$, $\text{Q}$, and $\text{R}$, respectively. Here, $A_1$, $A_2$, and $A_3$ represent the atomic masses of $\text{P}$, $\text{Q}$, and $\text{R}$, respectively.

The emission of a $\beta$ particle corresponds to the emission of an electron. The chemical reaction for $\beta$ emission from a parent nucleus to form a daughter nucleus is expressed as

$\text{P}\to \text{D}+e$

Here, $\text{P}$ is the parent nucleus, $A$ and $Z$ are the atomic mass and atomic number of the parent nucleus, respectively, $\text{D}$ is the daughter nucleus, $a$ and $b$ are the atomic mass and atomic number of daughter nucleus, respectively, and $e$ is the beta particle emitted.

Explanation:

Write a standard expression for the emission of abeta particle from the nucleus ${\text{UX}}_1$ to form the nucleus ${\text{UX}}_2$ as

$\text{U}{\text{X}}_1\to \text{U}{\text{X}}_2+e$

Here, $a$ and $b$ indicate the mass number and atomic number of the nucleus ${\text{UX}}_2$, respectively.

From part (a), the parent nucleus for this reaction is $\text{U}{\text{X}}_1$ and the daughter nucleus formed is ${\text{UX}}_2$. Therefore, substitute $234$ for $x$ and $90$ for $y$

$\text{U}{\text{X}}_1\to \text{U}{\text{X}}_2+e$ …… (2)

Apply the conservation of mass by equating the atomic masses on both sides of chemical equation (2):

$\begin{array}{c}234=a+0\\ a=234\end{array}$

Apply the conservation of mass by equating the atomic numbers on both sides of chemical equation (2):

$\begin{array}{c}90=b-1\\ b=90+1\\ b=91\end{array}$

The value of $a$ is calculated as $234$ and the value of $b$ is calculated as $91$.

Rewrite equation (2) as

$\text{U}{\text{X}}_1\to \text{U}{\text{X}}_2+e$

Substitute $234$ for $a$ and $90$ for $y$

$\text{U}{\text{X}}_1\to \text{U}{\text{X}}_2+e$

Conclusion:

The complete equation, after filling the values of $a$ and $b$, is $\text{U}{\text{X}}_1\to \text{U}{\text{X}}_2+e$. The daughter nucleus is represented as $\text{U}{\text{X}}_2$, which indicates that the atomic number of the daughter nucleus ${\text{UX}}_2$ is $91$ and the mass number of ${\text{UX}}_2$ is $234$.