Chapter 18, Problem 9E

Interpretation Introduction

(a)

Interpretation:

The equation for $\text{U}-238$ decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 9E

The equation for $\text{U}-238$ decay by alpha emission is shown below.

$\text{U}\to \text{T}\text{h}+\text{H}\text{e}$

Explanation of Solution

It is given that $\text{U-}238$ undergoes decay by alpha emission. An alpha particle is $\text{H}\text{e}$. Therefore, in the case of alpha decay the atomic number of an element decreases by $2$ and the mass number decreases by $4$. The atomic number of uranium is $92$ which decreases by $2$, that means resulting element has atomic number $90$. The element which has atomic number $90$ is thorium.

The resulting equation is shown below.

$\text{U}\to \text{T}\text{h}+\text{H}\text{e}$

Conclusion

The equation of $\text{U-}238$ decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for $\text{Al}-28$ decay by beta emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 9E

The equation for $\text{Al}-28$ decay by beta emission is shown below.

$\text{A}\text{l}\to \text{S}\text{i}+\text{e}$

Explanation of Solution

It is given that $\text{Al-}28$ undergoes decay by beta emission. A beta particle is identical to an electron that is $e$. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of aluminum is $13$ which increases by $1$ that means resulting element has atomic number $14$. The element which has an atomic number $14$ is silicon.

The resulting equation is shown below.

$\text{A}\text{l}\to \text{S}\text{i}+\text{e}$

Conclusion

The equation of $\text{Al-}28$ decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for $\text{O-}15$ decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 9E

The equation for $\text{O-}15$ decay by positron emission is shown below.

$\text{O}\to \text{N}+\text{e}$

Explanation of Solution

It is given that $\text{O-}15$ undergoes decay by positron emission. A positron is opposite of an electron and represented by $e$. Therefore, in the case of positron the atomic number of an element decreases by $1$ and there is no change in mass number. The atomic number of oxygen is $8$ which decreases by $1$ that means resulting element has atomic number $7$. The element which has an atomic number $7$ is nitrogen.

The resulting equation is shown below.

$\text{O}\to \text{N}+\text{e}$

Conclusion

The equation of $\text{O-}15$ decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for $\text{Fe-}55$ decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer to Problem 9E

The equation for $\text{Fe-}55$ decay by electron capture is shown below.

$\text{F}\text{e}+\text{e}\to \text{M}\text{n}$

Explanation of Solution

It is given that $\text{Fe}-55$ undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by $1$, and there is no change in mass number. The atomic number of iron is $26$ which decreases by $1$, that means resulting element has atomic number $25$. The element which has an atomic number $25$ is manganese.

The resulting equation is shown below.

$\text{F}\text{e}+\text{e}\to \text{M}\text{n}$

Conclusion

The equation of $\text{Fe}-55$ decay is stated above.