Chapter 25, Problem 41A

(a)

Interpretation Introduction

Interpretation:

The radionuclide formed by the decay of $\text{P}\text{u}$ by emitting alpha particles is to be determined.

Concept Interpretation:

A radioactive metal with a silvery-gray color and the molecular symbol $\text{Pu}$ , plutonium becomes yellow when it comes into contact with air.

Answer to Problem 41A

  $\text{U}$

Explanation of Solution

When a radioactive nucleus decays, the atomic nucleus emits an alpha particle (similar to that of a helium nucleus), after which the nucleus is transmuted into a different atomic number.

This is how an alpha decay's nuclear equation looks:

  $\text{X}\to \text{Y}\text{+}\text{H}\text{e}$

Since an alpha particle (helium nucleus) is emitted in alpha decay, the mass number $\left(\text{A}\right)$ of the nucleus $\text{X}$ is reduced by $\text{4}$ and the atomic number $\left(\text{Z}\right)$ is reduced by $2$ producing a different atomic nucleus $\text{Y}$

When Plutonium $\text{-238}$ experiences an alpha decay:

  $\text{P}\text{u}\to \text{U}\text{+}\text{H}\text{e}$

The end result is uranium $\text{-234}$ . $$

(b)

Interpretation Introduction

Interpretation:

  $\text{B}\text{i}$ is radioactive nuclei decay by emitting alpha particles. The new nuclei formed is to be determined.

Concept Interpretation:

  $\text{B}\text{i}$ can be handled as if it were non-radioactive for almost all applications due to its incredibly long half-life.

Answer to Problem 41A

  $\text{T}\text{i}$

Explanation of Solution

When a radioactive nucleus decays, the atomic nucleus emits an alpha particle (similar to that of a helium nucleus), which is followed by the nucleus changing into another atomic nucleus.

This is how an alpha decay's nuclear equation looks:

  $\text{X}\to \text{Y}\text{+}\text{H}\text{e}$

Since an alpha particle (helium nucleus) is emitted in alpha decay, the mass number $\left(\text{A}\right)$ of the nucleus $\text{X}$ is reduced by $\text{4}$ and the atomic number $\left(\text{Z}\right)$ is reduced by $2$ producing a different atomic nucleus $\text{Y}$

When Bismuth $\text{-210}$ experiences an alpha decay:

  $\text{B}\text{i}\to \text{T}\text{i+}\text{H}\text{e}$

The substance is thallium $\text{-206}$ .

(c)

Interpretation Introduction

Interpretation:

  $\text{P}\text{o}$ is radioactive nuclei decay by emitting alpha particles. The new nuclide formed is to be determined.

Concept Interpretation:

Industry uses polonium $\text{-210}$ to create anti-static devices. This helps produce things like tape, rolling paper, and spinning synthetic fibers.

Answer to Problem 41A

  $\text{P}\text{b}$

Explanation of Solution

When a radioactive nucleus decays, the atomic nucleus emits an alpha particle (similar to that of a helium nucleus), which is followed by the nucleus changing into another atomic nucleus.

This is how an alpha decay's nuclear equation looks:

  $\text{X}\to \text{Y}\text{+}\text{H}\text{e}$

Since an alpha particle (helium nucleus) is emitted in alpha decay, the mass number $\left(\text{A}\right)$ of the nucleus $\text{X}$ is reduced by $\text{4}$ and the atomic number $\left(\text{Z}\right)$ is reduced by $2$ producing a different atomic nucleus $\text{Y}$

When Polonium-210 experiences an alpha decay:

  $\text{P}\text{o}\to \text{P}\text{b+}\text{H}\text{e}$

Lead-206 is the product.

(d)

Interpretation Introduction

Interpretation:

  $\text{T}\text{h}$ is radioactive nuclei decay by emitting alpha particles. The new nuclide formed is to be determined.

Concept Interpretation:

Isotope thorium is used in welding rods and electric bulb filaments to improve product performance.

Answer to Problem 41A

  $\text{R}\text{a}$

Explanation of Solution

When a radioactive nucleus decays, the atomic nucleus emits an alpha particle (similar to that of a helium nucleus), which is followed by the nucleus changing into another atomic nucleus.

This is how an alpha decay's nuclear equation looks:

  $\text{X}\to \text{Y}\text{+}\text{H}\text{e}$

Since an alpha particle (helium nucleus) is emitted in alpha decay, the mass number $\left(\text{A}\right)$ of the nucleus $\text{X}$ is reduced by $\text{4}$ and the atomic number $\left(\text{Z}\right)$ is reduced by $2$ producing a different atomic nucleus $\text{Y}$

When thorium-230 experiences an alpha decay:

  $\text{T}\text{h}\to \text{R}\text{a+}\text{H}\text{e}$

Radium- $\text{226}$ is the product.