### Identifying the Missing Nuclides in Nuclear DecayThis exercise involves identifying the missing nuclides in several nuclear decay equations. Nuclear decay involves transformations that change one element into another. Below are the tasks and instructions for finding the unknown nuclides.#### Problem (a)**Decay Equation:**\[ \ce{^{212}_{83}Bi -> \underset{?}{X} + \ce{^{4}_{2}He}} \]**chemPad Interactive Pane:**- There is a text box where you will enter the symbol of the resulting nuclide after the decay.- The control panel includes options for script notation and Greek letters to assist in correctly formatting your answer.**Explanation:**- Bismuth-212 (\( \ce{^{212}_{83}Bi} \)) undergoes alpha decay, emitting a helium-4 nucleus (\( \ce{^{4}_{2}He} \)).- Use the conservation of mass and atomic numbers to find the resulting nuclide. - Atomic number: \( 83 - 2 = 81 \) - Mass number: \( 212 - 4 = 208 \)- The missing nuclide is \( \ce{^{208}_{81}Tl} \) (Thallium-208).#### Problem (b)**Decay Equation:**\[ \ce{^{95}_{36}Kr -> \underset{?}{X} + e^- + \bar{\nu}} \]**chemPad Interactive Pane:**- There is a text box where you will enter the symbol of the resulting nuclide.- The control panel helps in correctly formatting your answer.**Explanation:**- Krypton-95 (\( \ce{^{95}_{36}Kr} \)) undergoes beta-minus decay, emitting an electron (\( e^- \)) and an antineutrino (\( \bar{\nu} \)).- Use the conservation of mass and atomic numbers to find the resulting nuclide. - Atomic number: \( 36 + 1 = 37 \) - Mass number remains unchanged: \( 95 \)- The missing nuclide is \( \ce{^{95}_{37}Rb} \) (Rubidium-95).#### Problem (c)**Decay Equation:**\[ \underset{?}{\ce{X}} -> \ce{^{4}_{2

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Identify the missing nuclides in the following decays.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Concept explainers

Nuclear Fission

Nuclear fission is a type of nuclear reaction in which heavier nuclei split into lighter nuclei by releasing a large quantity of energy. The elements like uranium-235 and plutonium-239 isotope undergo nuclear fission releasing energy. When nuclei undergo fission, some quantity of mass is lost. The lost mass is converted to nuclear energy. Nuclear fission reaction is used in nuclear power plants and atomic bombs. In nuclear power plants, nuclear reactors are used to generate electricity.

Endothermic Nuclear Reaction

A nuclear reaction can be described as a process where two atoms, or two nuclei or nucleus and subatomic particles such as a proton, neutron interact together, and a large amount of energy is produced and new elements are also produced.

Question

### Identifying the Missing Nuclides in Nuclear Decay

This exercise involves identifying the missing nuclides in several nuclear decay equations. Nuclear decay involves transformations that change one element into another. Below are the tasks and instructions for finding the unknown nuclides.

#### Problem (a)

**Decay Equation:**
\[ \ce{^{212}_{83}Bi -> \underset{?}{X} + \ce{^{4}_{2}He}} \]

**chemPad Interactive Pane:**
- There is a text box where you will enter the symbol of the resulting nuclide after the decay.
- The control panel includes options for script notation and Greek letters to assist in correctly formatting your answer.

**Explanation:**
- Bismuth-212 (\( \ce{^{212}_{83}Bi} \)) undergoes alpha decay, emitting a helium-4 nucleus (\( \ce{^{4}_{2}He} \)).
- Use the conservation of mass and atomic numbers to find the resulting nuclide.
- Atomic number: \( 83 - 2 = 81 \)
- Mass number: \( 212 - 4 = 208 \)
- The missing nuclide is \( \ce{^{208}_{81}Tl} \) (Thallium-208).

#### Problem (b)

**Decay Equation:**
\[ \ce{^{95}_{36}Kr -> \underset{?}{X} + e^- + \bar{\nu}} \]

**chemPad Interactive Pane:**
- There is a text box where you will enter the symbol of the resulting nuclide.
- The control panel helps in correctly formatting your answer.

**Explanation:**
- Krypton-95 (\( \ce{^{95}_{36}Kr} \)) undergoes beta-minus decay, emitting an electron (\( e^- \)) and an antineutrino (\( \bar{\nu} \)).
- Use the conservation of mass and atomic numbers to find the resulting nuclide.
- Atomic number: \( 36 + 1 = 37 \)
- Mass number remains unchanged: \( 95 \)
- The missing nuclide is \( \ce{^{95}_{37}Rb} \) (Rubidium-95).

#### Problem (c)

**Decay Equation:**
\[ \underset{?}{\ce{X}} -> \ce{^{4}_{2

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