24.26 P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: ${}^{\mathbf{237}}\mathit{N}\mathit{p}$ is the parent nuclide of a decay series that starts with  $\mathbf{\alpha }$-emission, followed by   ${\mathbf{\beta }}^{\mathbf{-}}$ emission, and then two more emissions. Write a balanced nuclear equation for each step?

Step-by-Step Solution

Answer

Protactinium has atomic number 91, the product is protactinium-233, and the balanced reaction is 

 ${}_{93}{}^{237} \text{Np}{\to }_2^4\text{He}{+}_{91}^{233} \text{Pa}$

Uranium has the atomic number 92, the product is uranium- 233, and the balanced reaction is 

 ${}_{91}{}^{233} \text{Pa}{\to }_{-1}^0\text{e}{+}_{92}^{233}\text{U}$

Thorium has an atomic number of 90, the product is thorium-229, and the balanced reaction is;

${}_{92}{}^{233}\text{U}{\to }_2^4\text{He}{+}_{90}^{229}$

Radium is the element with atomic number 88, so the product is radium-225, and the balanced reaction is 

 ${}_{90}{}^{229}\text{Th}{\to }_2^4\text{He}{+}_{88}^{225}\text{Ra}$

Step 1: The given data

The parent nuclide's name is ${}^{237}Np$

The atomic weight of Np is 93.

The alpha particle ${}_{\text{2}}{}^{\text{4}}\text{He}$

 $\text{Electron: }{}_{-1}{}^{0}e$

Assume that the product nuclide is  Alpha decay of with atomic mass A and atomic number Z 

Let's compute the values of  

 $\begin{array}{rcl}& & \text{237 = 4 + A}\\ & & \text{A}\text{=237 - 4}\\ & & \text{ = 233}\\ & & \text{93 = 2 + Z}\\ & & \text{Z = 93 - 2}\\ & & \text{ = 91}\end{array}$

Because protactinium has atomic number 91, the product is protactinium-233, and the balanced reaction is 

 ${}_{93}{}^{237} \text{Np}{\to }_2^4\text{He}{+}_{91}^{233} \text{Pa}$

Step 2: Beta decay

The beta decay of ${}_{91}{}^{233} \text{Pa}$

 ${}_{91}{}^{233} \text{Pa}{\to }_{-1}^0\text{e}{+}_{\text{Z}}^{\text{A}}\text{X}$

Let's compute the values of A and Z.

 $$\begin{gathered} \text{233 = 0 + A} \\ \text{A = 233} \\ \text{91 = - 1 + Z} \\ \text{Z = 91 + 1} \\ \text{ = 92} \end{gathered}$$

Step 3: Balancing the Equation

Because uranium has the atomic number 92, the product is uranium- 233, and the balanced reaction is 

 ${}_{91}{}^{233} \text{Pa}{\to }_{-1}^0\text{e}{+}_{92}^{233}\text{U}$

The alpha decay of  ${}^{\text{233}}\text{U}$

 ${}_{92}{}^{233}\text{U}{\to }_2^4\text{He}{+}_{\text{Z}}^{\text{A}}\text{X}$

Let's compute the values of  A and Z

 $\begin{array}{rcl}& & \text{233 = 4 + A}\\ & & \text{A = 233 - 4 = 229}\\ & & \text{92 = 2 + Z}\\ & & \text{Z = 92 - 2}\\ & & \text{ = 90}\end{array}$

Because thorium has an atomic number of 90, the product is thorium-229, and the balanced reaction is

${}_{92}{}^{233}\text{U}{\to }_2^4\text{He}{+}_{90}^{229}$

Step 4: Alpha decay

The alpha decay of ${}^{\text{229}}\text{Th}$

 ${}_{90}{}^{229}\text{Th}{\to }_2^4\text{He}{+}_{\text{Z}}^{\text{A}}\text{X}$

Let's compute the values of A and Z

 $\begin{array}{rcl}& & \text{229 = 4 + A}\\ & & \text{A = 229 - 4}\\ & & \text{ = 225}\\ & & \text{90 = 2 + Z}\\ & & \text{Z = 90 - 2}\\ & & \text{ = 88}\end{array}$

Radium is the element with atomic number 88, so the product is radium-225, and the balanced reaction is 

 ${}_{90}{}^{229}\text{Th}{\to }_2^4\text{He}{+}_{88}^{225}\text{Ra}$

Hence, the overall equations are,

 $$\begin{gathered} {}_{93}^{237} \text{Np}{\to }_2^4\text{He}{+}_{91}^{233} \text{Pa}{ \\ }_{91}^{233} \text{Pa}{\to }_{-1}^0\text{e}{+}_{92}^{233}\text{U }{ \\ }_{92}^{233}\text{U}{\to }_2^4\text{He}{+}_{90}^{229}\text{Th}{ \\ }_{90}^{229}\text{Th}{\to }_2^4\text{He}{+}_{88}^{225}\text{Ra} \end{gathered}$$