Question

complete the following equations.
1.
2.
3.
4.
5.
6.
7.
8.

Explanation:

Step1: Determine the mass number and atomic number of the missing particle in each equation

For nuclear equations, the sum of mass numbers and atomic numbers on the left - hand side must be equal to the sum of mass numbers and atomic numbers on the right - hand side.

Equation 1: \(^{212}_{84}\text{Po}\to^{4}_{2}\text{He}+\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(212 = 4+A\), so \(A=212 - 4=208\)
For atomic number: \(84=2 + Z\), so \(Z = 84- 2=82\). The element with \(Z = 82\) is \(Pb\). So the missing particle is \(^{208}_{82}\text{Pb}\)

Equation 2: \(^{142}_{61}\text{Pm}+\to^{142}_{60}\text{Nd}\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(142+A = 142\), so \(A = 0\)
For atomic number: \(61+Z=60\), so \(Z=- 1\). The particle is \(^{\ \ 0}_{-1}\text{e}\) (beta - minus particle)

Equation 3: \(^{253}_{99}\text{Es}+^{4}_{2}\text{He}\to^{1}_{0}\text{n}+\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(253 + 4=1+A\), so \(A=253 + 4-1=256\)
For atomic number: \(99+2=0 + Z\), so \(Z = 101\). The element with \(Z = 101\) is \(Md\). So the missing particle is \(^{256}_{101}\text{Md}\)

Equation 4: \(^{218}_{84}\text{Po}\to^{4}_{2}\text{He}+\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(218=4+A\), so \(A=218 - 4=214\)
For atomic number: \(84=2 + Z\), so \(Z = 84- 2=82\). The element with \(Z = 82\) is \(Pb\). So the missing particle is \(^{214}_{82}\text{Pb}\)

Equation 5: \(^{9}_{4}\text{Be}+^{4}_{2}\text{He}\to+\ ^{1}_{0}\text{n}\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(9+4=A + 1\), so \(A=9 + 4-1=12\)
For atomic number: \(4+2=Z+0\), so \(Z = 6\). The element with \(Z = 6\) is \(C\). So the missing particle is \(^{12}_{6}\text{C}\)

Equation 6: \(^{22}_{11}\text{Na}+\to^{22}_{10}\text{Ne}\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(22+A = 22\), so \(A = 0\)
For atomic number: \(11+Z=10\), so \(Z=-1\). The particle is \(^{\ \ 0}_{-1}\text{e}\) (beta - minus particle)

Equation 7: \(^{238}_{92}\text{U}\to+\ ^{234}_{90}\text{Th}\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(238=A + 234\), so \(A=238 - 234=4\)
For atomic number: \(92=Z + 90\), so \(Z = 2\). The particle is \(^{4}_{2}\text{He}\)

Equation 8: \(^{37}\text{Ar}+\to^{37}_{17}\text{Cl}\)

Let the mass number of the missing particle be \(A\) and atomic number be \(Z\).
For mass number: \(37+A = 37\), so \(A = 0\)
For atomic number: \(18+Z=17\), so \(Z=-1\). The particle is \(^{\ \ 0}_{-1}\text{e}\) (beta - minus particle)

Answer:

1. \(^{208}_{82}\text{Pb}\)
2. \(^{\ \ 0}_{-1}\text{e}\)
3. \(^{256}_{101}\text{Md}\)
4. \(^{214}_{82}\text{Pb}\)
5. \(^{12}_{6}\text{C}\)
6. \(^{\ \ 0}_{-1}\text{e}\)
7. \(^{4}_{2}\text{He}\)
8. \(^{\ \ 0}_{-1}\text{e}\)