Question

1. (_{1}^{1}h+_{1}^{3}h

ightarrow)

1. (_{3}^{6}li+_{0}^{1}n

ightarrow_{2}^{4}he+)

1. (_{13}^{27}al+_{2}^{4}he

ightarrow_{15}^{30}p+)

1. (_{4}^{9}be+_{1}^{1}h

ightarrow) (+_{2}^{4}he)

1. (_{}^{37}k

ightarrow_{+1}^{0}e+)

1. (+ _{0}^{1}n

ightarrow_{56}^{142}ba+_{36}^{91}kr+3_{0}^{1}n)

1. (_{92}^{238}u+_{2}^{4}he

ightarrow) (+_{0}^{1}n)

Explanation:

Step1: Determine the mass number and atomic number for each equation

For nuclear reactions, the sum of mass numbers (superscripts) and atomic numbers (subscripts) on the left - hand side must equal the sum on the right - hand side.

Equation 4:

• Mass number: \(1 + 3=4\)
• Atomic number: \(1 + 1 = 2\)

The element with atomic number \(2\) is \(He\). So the product is \(_{2}^{4}He\)

Equation 5:

• Mass number: \(6+1 - 4=3\)
• Atomic number: \(3 + 0-2 = 1\)

The element with atomic number \(1\) is \(H\). So the product is \(_{1}^{3}H\)

Equation 6:

• Mass number: \(27 + 4-30=1\)
• Atomic number: \(13+2 - 15=0\)

The particle with mass number \(1\) and atomic number \(0\) is \(_{0}^{1}n\)

Equation 7:

• Mass number: \(9 + 1-4=6\)
• Atomic number: \(4+1 - 2=3\)

The element with atomic number \(3\) is \(Li\). So the product is \(_{3}^{6}Li\)

Equation 8:

• Mass number: \(37-0 = 37\)
• Atomic number: \(19-(+ 1)=18\)

The element with atomic number \(18\) is \(Ar\). So the product is \(_{18}^{37}Ar\)

Equation 9:

• Mass number: \(142+91 + 3-1=235\)
• Atomic number: \(56+36+0 - 0=92\)

The element with atomic number \(92\) is \(U\). So the reactant is \(_{92}^{235}U\)

Equation 10:

• Mass number: \(238+4 - 1=241\)
• Atomic number: \(92+2-0 = 94\)

The element with atomic number \(94\) is \(Pu\). So the product is \(_{94}^{241}Pu\)

Answer:

1. \(_{2}^{4}He\)
2. \(_{1}^{3}H\)
3. \(_{0}^{1}n\)
4. \(_{3}^{6}Li\)
5. \(_{18}^{37}Ar\)
6. \(_{92}^{235}U\)
7. \(_{94}^{241}Pu\)