Nuclear Decay Equations Worksheet for practicing alpha, beta, and positron decay reactions.

Problem Analysis:

The worksheet involves solving nuclear decay equations for three types of radioactive decay processes:
1. Alpha Decay: The nucleus emits an alpha particle (${}_{2}^{4}\text{He}$), which consists of 2 protons and 2 neutrons.
2. Beta Decay: The nucleus emits a beta particle (${}_{-1}^{0}\text{e}$), which is an electron. This process occurs when a neutron in the nucleus converts into a proton, emitting an electron and an antineutrino.
3. Positron Emission: The nucleus emits a positron (${}_{+1}^{0}\text{e}$), which is the antimatter counterpart of an electron. This process occurs when a proton in the nucleus converts into a neutron, emitting a positron and a neutrino.

For each type of decay, we need to determine the resulting nuclide by applying the conservation laws of mass number (A) and atomic number (Z).

---

Solution:

#### 1. Alpha Decay
In alpha decay, the mass number decreases by 4, and the atomic number decreases by 2. The general equation is:
\[
{}_{Z}^{A}\text{X} \rightarrow {}_{Z-2}^{A-4}\text{Y} + {}_{2}^{4}\text{He}
\]

##### Step-by-Step Solutions:
1. ${}_{84}^{210}\text{Po} \rightarrow \_\_\_\_ + {}_{2}^{4}\text{He}$
- Mass number: \(210 - 4 = 206\)
- Atomic number: \(84 - 2 = 82\)
- The element with atomic number 82 is Lead (Pb).
- Result: \({}_{84}^{210}\text{Po} \rightarrow {}_{82}^{206}\text{Pb} + {}_{2}^{4}\text{He}\)

2. ${}_{86}^{222}\text{Rn} \rightarrow \_\_\_\_ + {}_{2}^{4}\text{He}$
- Mass number: \(222 - 4 = 218\)
- Atomic number: \(86 - 2 = 84\)
- The element with atomic number 84 is Polonium (Po).
- Result: \({}_{86}^{222}\text{Rn} \rightarrow {}_{84}^{218}\text{Po} + {}_{2}^{4}\text{He}\)

3. ${}_{90}^{238}\text{Th} \rightarrow \_\_\_\_ + {}_{2}^{4}\text{He}$
- Mass number: \(238 - 4 = 234\)
- Atomic number: \(90 - 2 = 88\)
- The element with atomic number 88 is Radon (Rn).
- Result: \({}_{90}^{238}\text{Th} \rightarrow {}_{88}^{234}\text{Rn} + {}_{2}^{4}\text{He}\)

4. ${}_{92}^{238}\text{U} \rightarrow \_\_\_\_ + {}_{2}^{4}\text{He}$
- Mass number: \(238 - 4 = 234\)
- Atomic number: \(92 - 2 = 90\)
- The element with atomic number 90 is Thorium (Th).
- Result: \({}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He}\)

5. ${}_{94}^{239}\text{Pu} \rightarrow \_\_\_\_ + {}_{2}^{4}\text{He}$
- Mass number: \(239 - 4 = 235\)
- Atomic number: \(94 - 2 = 92\)
- The element with atomic number 92 is Uranium (U).
- Result: \({}_{94}^{239}\text{Pu} \rightarrow {}_{92}^{235}\text{U} + {}_{2}^{4}\text{He}\)

---

#### 2. Beta Decay
In beta decay, the mass number remains unchanged, but the atomic number increases by 1. The general equation is:
\[
{}_{Z}^{A}\text{X} \rightarrow {}_{Z+1}^{A}\text{Y} + {}_{-1}^{0}\text{e}
\]

##### Step-by-Step Solutions:
1. ${}_{6}^{14}\text{C} \rightarrow \_\_\_\_ + {}_{-1}^{0}\text{e}$
- Mass number: \(14\) (unchanged)
- Atomic number: \(6 + 1 = 7\)
- The element with atomic number 7 is Nitrogen (N).
- Result: \({}_{6}^{14}\text{C} \rightarrow {}_{7}^{14}\text{N} + {}_{-1}^{0}\text{e}\)

2. ${}_{19}^{40}\text{K} \rightarrow \_\_\_\_ + {}_{-1}^{0}\text{e}$
- Mass number: \(40\) (unchanged)
- Atomic number: \(19 + 1 = 20\)
- The element with atomic number 20 is Calcium (Ca).
- Result: \({}_{19}^{40}\text{K} \rightarrow {}_{20}^{40}\text{Ca} + {}_{-1}^{0}\text{e}\)

3. ${}_{7}^{13}\text{N} \rightarrow \_\_\_\_ + {}_{-1}^{0}\text{e}$
- Mass number: \(13\) (unchanged)
- Atomic number: \(7 + 1 = 8\)
- The element with atomic number 8 is Oxygen (O).
- Result: \({}_{7}^{13}\text{N} \rightarrow {}_{8}^{13}\text{O} + {}_{-1}^{0}\text{e}\)

4. ${}_{38}^{90}\text{Sr} \rightarrow \_\_\_\_ + {}_{-1}^{0}\text{e}$
- Mass number: \(90\) (unchanged)
- Atomic number: \(38 + 1 = 39\)
- The element with atomic number 39 is Yttrium (Y).
- Result: \({}_{38}^{90}\text{Sr} \rightarrow {}_{39}^{90}\text{Y} + {}_{-1}^{0}\text{e}\)

5. ${}_{46}^{105}\text{Pd} \rightarrow \_\_\_\_ + {}_{-1}^{0}\text{e}$
- Mass number: \(105\) (unchanged)
- Atomic number: \(46 + 1 = 47\)
- The element with atomic number 47 is Silver (Ag).
- Result: \({}_{46}^{105}\text{Pd} \rightarrow {}_{47}^{105}\text{Ag} + {}_{-1}^{0}\text{e}\)

---

#### 3. Positron Emission
In positron emission, the mass number remains unchanged, but the atomic number decreases by 1. The general equation is:
\[
{}_{Z}^{A}\text{X} \rightarrow {}_{Z-1}^{A}\text{Y} + {}_{+1}^{0}\text{e}
\]

##### Step-by-Step Solutions:
1. ${}_{29}^{61}\text{Cu} \rightarrow \_\_\_\_ + {}_{+1}^{0}\text{e}$
- Mass number: \(61\) (unchanged)
- Atomic number: \(29 - 1 = 28\)
- The element with atomic number 28 is Nickel (Ni).
- Result: \({}_{29}^{61}\text{Cu} \rightarrow {}_{28}^{61}\text{Ni} + {}_{+1}^{0}\text{e}\)

2. ${}_{9}^{18}\text{F} \rightarrow \_\_\_\_ + {}_{+1}^{0}\text{e}$
- Mass number: \(18\) (unchanged)
- Atomic number: \(9 - 1 = 8\)
- The element with atomic number 8 is Oxygen (O).
- Result: \({}_{9}^{18}\text{F} \rightarrow {}_{8}^{18}\text{O} + {}_{+1}^{0}\text{e}\)

3. ${}_{16}^{30}\text{S} \rightarrow \_\_\_\_ + {}_{+1}^{0}\text{e}$
- Mass number: \(30\) (unchanged)
- Atomic number: \(16 - 1 = 15\)
- The element with atomic number 15 is Phosphorus (P).
- Result: \({}_{16}^{30}\text{S} \rightarrow {}_{15}^{30}\text{P} + {}_{+1}^{0}\text{e}\)

4. ${}_{50}^{116}\text{Sn} \rightarrow \_\_\_\_ + {}_{+1}^{0}\text{e}$
- Mass number: \(116\) (unchanged)
- Atomic number: \(50 - 1 = 49\)
- The element with atomic number 49 is Indium (In).
- Result: \({}_{50}^{116}\text{Sn} \rightarrow {}_{49}^{116}\text{In} + {}_{+1}^{0}\text{e}\)

5. ${}_{21}^{40}\text{Sc} \rightarrow \_\_\_\_ + {}_{+1}^{0}\text{e}$
- Mass number: \(40\) (unchanged)
- Atomic number: \(21 - 1 = 20\)
- The element with atomic number 20 is Calcium (Ca).
- Result: \({}_{21}^{40}\text{Sc} \rightarrow {}_{20}^{40}\text{Ca} + {}_{+1}^{0}\text{e}\)

---

Final Answers:

1. Alpha Decay:
\[
\boxed{
\begin{aligned}
&{}_{84}^{210}\text{Po} \rightarrow {}_{82}^{206}\text{Pb} + {}_{2}^{4}\text{He} \\
&{}_{86}^{222}\text{Rn} \rightarrow {}_{84}^{218}\text{Po} + {}_{2}^{4}\text{He} \\
&{}_{90}^{238}\text{Th} \rightarrow {}_{88}^{234}\text{Rn} + {}_{2}^{4}\text{He} \\
&{}_{92}^{238}\text{U} \rightarrow {}_{90}^{234}\text{Th} + {}_{2}^{4}\text{He} \\
&{}_{94}^{239}\text{Pu} \rightarrow {}_{92}^{235}\text{U} + {}_{2}^{4}\text{He}
\end{aligned}
}
\]

2. Beta Decay:
\[
\boxed{
\begin{aligned}
&{}_{6}^{14}\text{C} \rightarrow {}_{7}^{14}\text{N} + {}_{-1}^{0}\text{e} \\
&{}_{19}^{40}\text{K} \rightarrow {}_{20}^{40}\text{Ca} + {}_{-1}^{0}\text{e} \\
&{}_{7}^{13}\text{N} \rightarrow {}_{8}^{13}\text{O} + {}_{-1}^{0}\text{e} \\
&{}_{38}^{90}\text{Sr} \rightarrow {}_{39}^{90}\text{Y} + {}_{-1}^{0}\text{e} \\
&{}_{46}^{105}\text{Pd} \rightarrow {}_{47}^{105}\text{Ag} + {}_{-1}^{0}\text{e}
\end{aligned}
}
\]

3. Positron Emission:
\[
\boxed{
\begin{aligned}
&{}_{29}^{61}\text{Cu} \rightarrow {}_{28}^{61}\text{Ni} + {}_{+1}^{0}\text{e} \\
&{}_{9}^{18}\text{F} \rightarrow {}_{8}^{18}\text{O} + {}_{+1}^{0}\text{e} \\
&{}_{16}^{30}\text{S} \rightarrow {}_{15}^{30}\text{P} + {}_{+1}^{0}\text{e} \\
&{}_{50}^{116}\text{Sn} \rightarrow {}_{49}^{116}\text{In} + {}_{+1}^{0}\text{e} \\
&{}_{21}^{40}\text{Sc} \rightarrow {}_{20}^{40}\text{Ca} + {}_{+1}^{0}\text{e}
\end{aligned}
}
\]