Question

Complete the following table with the nuclear particle that is produced in each nuclear reaction.

Short answer

The nuclear particle that is produced in each nuclear reaction is given below:

a) alpha particle ${}_2{}^4\mathrm{He}$,

b) beta particle ${}_{-1}{}^{0}e$,

c) beta particle ${}_{-1}{}^{0}e$,

d) beta particle ${}_{-1}{}^{0}e$,

e) beta particle ${}_{-1}{}^{0}e$,

Step-by-step solution

Types of particles in radioactivity

• A positively charged alpha particle is made up of two protons and two neutrons.
• A beta particle, on the other hand, is a charged particle with either a unit negative or positive charge.
• A gamma particle, on the other hand, has no charge and is therefore neutral.

Step 2: Particle produced by first nuclide.

The reaction is ${}_{94}{}^{239}\mathrm{Pu}{\to }_{92}^{235}\mathrm{U}{+}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}$:

Now the value of mass number (A) and atomic number (Z) is evaluated.

$\begin{array}{c}239=235+\mathrm{A}\\ \mathrm{A}=239-235\\ =4\\ \\ 94=92+\mathrm{Z}\\ \mathrm{Z}=94-92\\ =2\end{array}$

Mass number of the particle is 4.

Atomic number of particles is 2.

So, it is an alpha particle, ${}_2{}^4\mathrm{He}$.

Step 3: Particle produced by second nuclide.

The reaction is ${}_{82}{}^{214}\mathrm{Pb}{\to }_{83}^{214}\mathrm{Bi}{+}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}$:

Now the value of mass number (A) and atomic number (Z) is evaluated.

$\begin{array}{c}214=214+\mathrm{A}\\ \mathrm{A}=214-214\\ =0\\ \\ 82=83+\mathrm{Z}\\ \mathrm{Z}=82-83\\ =-1\end{array}$

Mass number of the particle is 0.

Atomic number of particles is -1.

So, it is beta particle, ${}_{-1}{}^{0}e$.

Step 4: Particle produced by third nuclide.

The reaction is:${}_{27}{}^{60}\mathrm{Co}{\to }_{28}^{60}\mathrm{Ni}{+}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}$

Now the value of mass number (A) and atomic number (Z) is evaluated.

$\begin{array}{c}60=60+\mathrm{A}\\ \mathrm{A}=60-60\\ =0\\ \\ 27=28+\mathrm{Z}\\ \mathrm{Z}=27-28\\ =-1\end{array}$

Mass number of the particle is 0.

Atomic number of particles is -1.

So, it is beta particle, ${}_{-1}{}^{0}e$.

Step 5: Particle produced by fourth nuclide

The reaction is:${}_{43}{}^{99}\mathrm{Tc}{\to }_{44}^{99}\mathrm{Ru}{+}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}$

Now the value of mass number (A) and atomic number (Z) is evaluated.

$\begin{array}{c}99=99+\mathrm{A}\\ \mathrm{A}=99-99\\ =0\\ \\ 43=44+\mathrm{Z}\\ \mathrm{Z}=43-44\\ =-1\end{array}$

Mass number of the particle is 0.

Atomic number of particles is -1.

So, it is beta particle, ${}_{-1}{}^{0}e$.

Particle produced by fifth nuclide

The reaction is:${}_{93}{}^{239}\mathrm{Np}{\to }_{94}^{239}\mathrm{Pu}{+}_{\mathrm{Z}}^{\mathrm{A}}\mathrm{X}$

Now the value of mass number (A) and atomic number (Z) is evaluated.

$\begin{array}{c}239=239+\mathrm{A}\\ \mathrm{A}=239-239\\ =0\\ \\ 93=94+\mathrm{Z}\\ \mathrm{Z}=93-94\\ =-1\end{array}$

Mass number of the particle is 0.

Atomic number of particles is -1.

So, it is beta particle, ${}_{-1}{}^{0}e{}_{-1}{}^{0}e$.