Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of $\text{U}-238$ through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than $83$ are radioactive.

Answer to Problem 11ST

Alpha emission of $\text{U}-238$ releases ${\text{Th}}_{90}^{234}$ and ${\text{He}}_{\text{2}}^{\text{4}}$ as shown below.

${}_{92}^{238}\text{U}{\to }_{90}^{234}\text{Th}{+}_2^4\text{He}$.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by $4$ and atomic number decreasesby $2$. The radioactive decay of $\text{U}-238$ by alpha emission releases one alpha particle ${\text{He}}_{\text{2}}^{\text{4}}$ and gives ${\text{Th}}_{90}^{234}$ as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of $\text{U}-238$ is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of $\text{U}-239$ through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than $83$ are radioactive.

Answer to Problem 11ST

Beta emission of $\text{U}-239$ releases ${}_{93}^{239}\text{Np}$ and ${\text{e}}_{-1}^0$ as shown below.

${}_{92}^{239}\text{U}{\to }_{93}^{239}\text{Np}{+}_{-1}^0\text{e}$

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by $1$ and mass number remains the same. When $\text{U}-239$ beta emission takes place, it releases one electron and gives $1$ increased atomic number element that is ${}_{93}^{239}\text{Np}$ as shpown in Figure 2.

Figure 2

Conclusion

Radioactive beta decay of $\text{U}-239$ is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of $\text{K}-40$ through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than $83$ are radioactive.

Answer to Problem 11ST

Positron emission of $\text{K}-40$ releases ${}_{18}^{40}\text{Ar}$ and ${\text{e}}_1^0$ as shown below.

${}_{19}^{40}\text{K}{\to }_{18}^{40}\text{Ar}{+}_{+1}^0\text{e}$

Explanation of Solution

In positron emission, an electron is released and its charge is $+1$ rather than $-1$. It can also be defined as the atomic number decreases by $1$ and mass number remains the same in Positron emission. The positron emission of $\text{K}-40$ is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of $\text{K}-40$ is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of $\text{K}-40$ through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than $83$ are radioactive.

Answer to Problem 11ST

Electron capture of $\text{K}-40$ releases ${}_{18}^{40}\text{Ar}$ as shown below.

${}_{19}^{40}\text{K}{+}_{-1}^0\text{e}{\to }_{18}^{40}\text{Ar}$

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of $\text{K}-40$ is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of $\text{K}-40$ is shown in Figure 4.