Chapter 19, Problem 19.10VP

(a)

Interpretation Introduction

Interpretation: The based on the given images are to be answered.

Concept introduction: The beta particle is a negatively charged particle with zero mass charge.

Fusion reaction is the one in which the number of reactants are more than the number of products.

The positron particle is a positively charged particle with zero mass charge.

Fission reaction is the one in which the number of reactants are less than the number of products.

$\text{Radon}-222$ and $\text{americium}-241$ undergo alpha decay.

$\text{Lead}-210$ and $\text{iodine}-131$ undergo beta decay.

To determine: The image that represents beta decay and a balanced nuclear equation to represent the image.

Answer to Problem 19.10VP

Solution:

Image $\text{B}$ represents beta decay as,

${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

Explanation of Solution

The beta particle is a negatively charged particle with zero mass charge.

In beta decay, a radioactive substance decomposes to emit a beta particle along with a nuclide with the atomic number more than that of the radioactive nuclide.

From the given images, image $\text{B}$ represents a nuclear reaction in which one substance dissociate into another substance with the removal of a negatively charge particle.

Figure 1

Therefore, image $\text{B}$ represents beta decay as,

${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

(b)

Interpretation Introduction

To determine: The image that represents fusion and a balanced nuclear equation to represent the image.

Answer to Problem 19.10VP

Solution:

Image $\text{D}$ represents fusion as,

${}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{\stackrel{}{\to }}_2^4\text{He}{+}_{-1}^0\text{β}{+}_{-1}^0\text{β}+\text{Energy}$

Explanation of Solution

Fusion reaction is the one in which the number of reactants are more than the number of products.

From the given images, image $\text{D}$ represents a nuclear reaction in which the number of reactants is four and the number of products is three with the release of energy.

If represents the fusion of four hydrogen atoms to form a helium atom and two beta particles.

Figure 2

Therefore, image $\text{D}$ represents fusion as,

${}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{\stackrel{}{\to }}_2^4\text{He}{+}_{-1}^0\text{β}{+}_{-1}^0\text{β}+\text{Energy}$

(c)

Interpretation Introduction

To determine: The image that represents positron emission and a balanced nuclear equation to represent the image.

Answer to Problem 19.10VP

Solution:

Image $\text{B}$ represents positron emission as,

${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

Explanation of Solution

The positron particle is a positively charged particle with zero mass charge.

In positron emission, a radioactive substance decomposes to emit a positron particle along with a nuclide with the atomic number less than that of the radioactive nuclide.

From the given images, image $\text{H}$ represents a nuclear reaction in which one substance dissociate into another substance with the removal of a positively charge particle.

Figure 3

Therefore, image $\text{H}$ represents positron emission as,

${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

(d)

Interpretation Introduction

To determine: The image that represents fission and a balanced nuclear equation to represent the image.

Answer to Problem 19.10VP

Solution:

Image $\text{F}$ represents fission as,

${}_{92}^{233}\text{U}{+}_0^1\text{n}{\stackrel{}{\to }}_{54}^{137}\text{Xe}{+}_{38}^{94}\text{Sr}+3_0^1\text{n}$

Explanation of Solution

Fission reaction is the one in which the number of reactants are less than the number of products.

From the given images, image $\text{F}$ represents a nuclear reaction in which two reactants break into six particles.

If represents the fission of $\text{uranium}-233$ reacting with neutron to form a xenon isotope, strontium isotope and three neutrons.

Figure 4

Therefore, image $\text{F}$ represents fission as,

${}_{92}^{233}\text{U}{+}_0^1\text{n}{\stackrel{}{\to }}_{54}^{137}\text{Xe}{+}_{38}^{94}\text{Sr}+3_0^1\text{n}$

(e)

Interpretation Introduction

To determine: The image that represents a nuclide that undergoes alpha decay and a balanced nuclear equation to represent the alpha decay of the nuclide.

Answer to Problem 19.10VP

Solution:

Image $\text{A}$ and image $\text{I}$ represents the nuclides that undergo alpha decay.

$\text{Radon}-222$ undergoes alpha decay as,

${}_{86}^{222}\text{Rn}{\stackrel{}{\to }}_{84}^{218}\text{Po}{+}_2^4\text{α}$

$\text{Americium}-241$ undergoes alpha decay as,

${}_{95}^{241}\text{Am}{\stackrel{}{\to }}_{93}^{237}\text{Np}{+}_2^4\text{α}+\text{γ}$

Explanation of Solution

Out of the given nuclides, the one that undergoes alpha decay are $\text{radon}-222$ and $\text{americium}-241$.

Image $\text{A}$ represents $\text{radon}-222$ and image $\text{I}$ represents $\text{americium}-241$.

Figure 5

Figure 6

Thus, image $\text{A}$ and image $\text{I}$ represents the nuclides that undergo alpha decay.

In alpha decay, the mass number of the radionuclide reduces by four and the atomic number reduces by two.

Therefore, $\text{radon}-222$ undergoes alpha decay as,

${}_{86}^{222}\text{Rn}{\stackrel{}{\to }}_{84}^{218}\text{Po}{+}_2^4\text{α}$

$\text{Americium}-241$ undergoes alpha decay as,

${}_{95}^{241}\text{Am}{\stackrel{}{\to }}_{93}^{237}\text{Np}{+}_2^4\text{α}+\text{γ}$

(f)

Interpretation Introduction

To determine: The image that represents a nuclide that undergoes beta decay and a balanced nuclear equation to represent the beta decay of the nuclide.

Answer to Problem 19.10VP

Solution:

Image $\text{C}$ and image $\text{G}$ represents the nuclides that undergo beta decay.

$\text{Lead}-210$ and undergoes beta decay as,

${}_{82}^{210}\text{Pb}{\stackrel{}{\to }}_{83}^{210}\text{Bi}{+}_{-1}^0\text{β}$

$\text{Iodine}-131$ undergoes beta decay as,

${}_{53}^{131}\text{I}{\stackrel{}{\to }}_{54}^{131}\text{Xe}{+}_{-1}^0\text{β}$

Explanation of Solution

Out of the given nuclides, the one that undergoes beta decay are $\text{lead}-210$ and $\text{iodine}-131$.

Image $\text{C}$ represents $\text{lead}-210$ and image $\text{G}$ represents $\text{iodine}-131$.

Figure 7

Figure 8

Thus, image $\text{C}$ and image $\text{G}$ represents the nuclides that undergo beta decay.

In beta decay, the mass number of the radionuclide remains same and the atomic number increases by one.

Therefore, $\text{lead}-210$ undergoes beta decay as,

${}_{82}^{210}\text{Pb}{\stackrel{}{\to }}_{83}^{210}\text{Bi}{+}_{-1}^0\text{β}$

$\text{Iodine}-131$ undergoes beta decay as,

${}_{53}^{131}\text{I}{\stackrel{}{\to }}_{54}^{131}\text{Xe}{+}_{-1}^0\text{β}$

Conclusion:

1. a. Image $\text{B}$ represents beta decay as,

   ${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

2. b. Image $\text{D}$ represents fusion as,

   ${}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{+}_1^2\text{H}{\stackrel{}{\to }}_2^4\text{He}{+}_{-1}^0\text{β}{+}_{-1}^0\text{β}+\text{Energy}$

3. c. Image $\text{B}$ represents positron emission as,

   ${}_Z^A\text{X}{\stackrel{}{\to }}_{Z+1}^A{\text{Y+}}_{-1}^0\text{β}$

4. d. Image $\text{F}$ represents fission as,

   ${}_{92}^{233}\text{U}{+}_0^1\text{n}{\stackrel{}{\to }}_{54}^{137}\text{Xe}{+}_{38}^{94}\text{Sr}+3_0^1\text{n}$

5. e. Image $\text{A}$ and image $\text{I}$ represents the nuclides that undergo alpha decay.

   $\text{Radon}-222$ undergoes alpha decay as,

   ${}_{86}^{222}\text{Rn}{\stackrel{}{\to }}_{84}^{218}\text{Po}{+}_2^4\text{α}$

   $\text{Americium}-241$ undergoes alpha decay as,

   ${}_{95}^{241}\text{Am}{\stackrel{}{\to }}_{93}^{237}\text{Np}{+}_2^4\text{α}+\text{γ}$

6. f. Image $\text{C}$ and image $\text{G}$ represents the nuclides that undergo beta decay.

   $\text{Lead}-210$ and undergoes beta decay as,

   ${}_{82}^{210}\text{Pb}{\stackrel{}{\to }}_{83}^{210}\text{Bi}{+}_{-1}^0\text{β}$

   $\text{Iodine}-131$ undergoes beta decay as,

   ${}_{53}^{131}\text{I}{\stackrel{}{\to }}_{54}^{131}\text{Xe}{+}_{-1}^0\text{β}$