Worksheet for practicing nuclear equations and decay processes.

Problem Analysis:

The task involves completing nuclear equations for various types of radioactive decay (beta decay, alpha decay) and answering questions related to nuclear reactions. Let's break it down step by step.

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Part 1: Complete the Following Nuclear Equations

#### Beta Decay
In beta decay, a neutron in the nucleus is converted into a proton, emitting an electron (β⁻) and an antineutrino (not shown in the equation). The atomic number increases by 1, while the mass number remains the same.

#### Alpha Decay
In alpha decay, the nucleus emits an alpha particle (⁴He), which consists of 2 protons and 2 neutrons. The atomic number decreases by 2, and the mass number decreases by 4.

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#### Solutions for Beta Decay (Questions 32–35):

1. 32) Beta Decay: $_{26}^{59}\text{Fe} \rightarrow$
- Fe (Iron) has atomic number 26.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 27, which is Co (Cobalt).
- The mass number remains the same (59).
- Equation: $_{26}^{59}\text{Fe} \rightarrow _{27}^{59}\text{Co} + _{-1}^{0}\text{e}$

2. 33) Beta Decay: $_{92}^{238}\text{U} \rightarrow$
- U (Uranium) has atomic number 92.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 93, which is Np (Neptunium).
- The mass number remains the same (238).
- Equation: $_{92}^{238}\text{U} \rightarrow _{93}^{238}\text{Np} + _{-1}^{0}\text{e}$

3. 34) Beta Decay: $_{19}^{40}\text{K} \rightarrow$
- K (Potassium) has atomic number 19.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 20, which is Ca (Calcium).
- The mass number remains the same (40).
- Equation: $_{19}^{40}\text{K} \rightarrow _{20}^{40}\text{Ca} + _{-1}^{0}\text{e}$

4. 35) Beta Decay: $_{90}^{234}\text{Th} \rightarrow$
- Th (Thorium) has atomic number 90.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 91, which is Pa (Protactinium).
- The mass number remains the same (234).
- Equation: $_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e}$

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#### Solutions for Alpha Decay (Questions 36–39):

1. 36) Alpha Decay: $_{88}^{226}\text{Ra} \rightarrow$
- Ra (Radium) has atomic number 88.
- In alpha decay, the atomic number decreases by 2, so the new element will have atomic number 86, which is Rn (Radon).
- The mass number decreases by 4, so the new mass number is 222.
- Equation: $_{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He}$

2. 37) Beta Decay: $_{95}^{241}\text{Am} \rightarrow$
- Am (Americium) has atomic number 95.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 96, which is Cm (Curium).
- The mass number remains the same (241).
- Equation: $_{95}^{241}\text{Am} \rightarrow _{96}^{241}\text{Cm} + _{-1}^{0}\text{e}$

3. 38) Alpha Decay: $_{29}^{64}\text{Cu} \rightarrow$
- Cu (Copper) has atomic number 29.
- In alpha decay, the atomic number decreases by 2, so the new element will have atomic number 27, which is Co (Cobalt).
- The mass number decreases by 4, so the new mass number is 60.
- Equation: $_{29}^{64}\text{Cu} \rightarrow _{27}^{60}\text{Co} + _{2}^{4}\text{He}$

4. 39) Beta Decay: $_{84}^{210}\text{Po} \rightarrow$
- Po (Polonium) has atomic number 84.
- In beta decay, the atomic number increases by 1, so the new element will have atomic number 85, which is At (Astatine).
- The mass number remains the same (210).
- Equation: $_{84}^{210}\text{Po} \rightarrow _{85}^{210}\text{At} + _{-1}^{0}\text{e}$

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Part 2: Write the Particle that Completes Each Equation Below

#### Solutions:

1. 40) $_{84}^{210}\text{Po} \rightarrow _{83}^{209}\text{Tl} +$
- The atomic number decreases by 1, and the mass number decreases by 1. This indicates the emission of a β⁻ particle.
- Particle: $_{-1}^{0}\text{e}$

2. 41) $_{6}^{14}\text{C} \rightarrow _{7}^{14}\text{N} +$
- The atomic number increases by 1, and the mass number remains the same. This indicates the emission of a β⁻ particle.
- Particle: $_{-1}^{0}\text{e}$

3. 42) $_{56}^{137}\text{Ba} \rightarrow _{57}^{137}\text{La} +$
- The atomic number increases by 1, and the mass number remains the same. This indicates the emission of a β⁻ particle.
- Particle: $_{-1}^{0}\text{e}$

4. 43) $_{83}^{210}\text{Bi} \rightarrow _{84}^{210}\text{Po} +$
- The atomic number increases by 1, and the mass number remains the same. This indicates the emission of a β⁻ particle.
- Particle: $_{-1}^{0}\text{e}$

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Part 3: Use the Nuclear Equation Below to Answer the Following Questions

The given nuclear equation is:
$$
_{84}^{210}\text{Po} \rightarrow _{2}^{4}\text{He} + _{82}^{206}\text{Pb}
$$

#### Questions:

1. 44) What picture above represents alpha nuclear reactions?
- The picture shows the emission of an alpha particle (⁴He), which is characteristic of alpha decay.
- Answer: Alpha decay.

2. 45) What is the mass of the starting Polonium (Po) atom?
- The mass number of the starting Polonium atom is 210.
- Answer: 210.

3. 46) What is the mass of the alpha particle?
- The alpha particle is $_{2}^{4}\text{He}$, so its mass is 4.
- Answer: 4.

4. 47) If an atom has too many protons, what type of decay will it likely go through? (ALPHA / BETA)
- An atom with too many protons is unstable and will likely undergo alpha decay to reduce the number of protons and neutrons.
- Answer: ALPHA.

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Final Answers:

1. 32) $_{26}^{59}\text{Fe} \rightarrow _{27}^{59}\text{Co} + _{-1}^{0}\text{e}$
2. 33) $_{92}^{238}\text{U} \rightarrow _{93}^{238}\text{Np} + _{-1}^{0}\text{e}$
3. 34) $_{19}^{40}\text{K} \rightarrow _{20}^{40}\text{Ca} + _{-1}^{0}\text{e}$
4. 35) $_{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e}$
5. 36) $_{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He}$
6. 37) $_{95}^{241}\text{Am} \rightarrow _{96}^{241}\text{Cm} + _{-1}^{0}\text{e}$
7. 38) $_{29}^{64}\text{Cu} \rightarrow _{27}^{60}\text{Co} + _{2}^{4}\text{He}$
8. 39) $_{84}^{210}\text{Po} \rightarrow _{85}^{210}\text{At} + _{-1}^{0}\text{e}$
9. 40) $_{-1}^{0}\text{e}$
10. 41) $_{-1}^{0}\text{e}$
11. 42) $_{-1}^{0}\text{e}$
12. 43) $_{-1}^{0}\text{e}$
13. 44) Alpha decay
14. 45) 210
15. 46) 4
16. 47) ALPHA

Final Boxed Answers:
\[
\boxed{
\begin{aligned}
&\text{32) } _{26}^{59}\text{Fe} \rightarrow _{27}^{59}\text{Co} + _{-1}^{0}\text{e} \\
&\text{33) } _{92}^{238}\text{U} \rightarrow _{93}^{238}\text{Np} + _{-1}^{0}\text{e} \\
&\text{34) } _{19}^{40}\text{K} \rightarrow _{20}^{40}\text{Ca} + _{-1}^{0}\text{e} \\
&\text{35) } _{90}^{234}\text{Th} \rightarrow _{91}^{234}\text{Pa} + _{-1}^{0}\text{e} \\
&\text{36) } _{88}^{226}\text{Ra} \rightarrow _{86}^{222}\text{Rn} + _{2}^{4}\text{He} \\
&\text{37) } _{95}^{241}\text{Am} \rightarrow _{96}^{241}\text{Cm} + _{-1}^{0}\text{e} \\
&\text{38) } _{29}^{64}\text{Cu} \rightarrow _{27}^{60}\text{Co} + _{2}^{4}\text{He} \\
&\text{39) } _{84}^{210}\text{Po} \rightarrow _{85}^{210}\text{At} + _{-1}^{0}\text{e} \\
&\text{40) } _{-1}^{0}\text{e} \\
&\text{41) } _{-1}^{0}\text{e} \\
&\text{42) } _{-1}^{0}\text{e} \\
&\text{43) } _{-1}^{0}\text{e} \\
&\text{44) Alpha decay} \\
&\text{45) 210} \\
&\text{46) 4} \\
&\text{47) ALPHA}
\end{aligned}
}
\]