Question

1. $^{42}k \to _{-1}^{0}e + \underline{}$
2. $^{239}pu \to _{2}^{4}he + \underline{}$
3. $_{92}^{235}u \to \underline{} + _{90}^{231}th$
4. $_{1}^{1}h + _{1}^{3}h \to \underline{}$
5. $_{3}^{6}li + _{0}^{1}n \to _{2}^{4}he + \underline{}$
6. $_{13}^{27}al + _{2}^{4}he \to _{15}^{30}p + \underline{}$
7. $_{4}^{9}be + _{1}^{1}h \to \underline{} + _{2}^{4}he$
8. $^{37}k \to _{+1}^{0}e + \underline{}$

Explanation:

Step1: Determine the missing particle in nuclear reactions

For nuclear reactions, we use the conservation of mass - number (A) and atomic - number (Z).

Problem 1: \(^{42}_{19}K

ightarrow^{0}_{- 1}e+\)

• Mass - number: \(A = 42-0=42\)
• Atomic - number: \(Z=19 - (-1)=20\)
• The element with \(Z = 20\) is calcium (\(Ca\)). So the particle is \(^{42}_{20}Ca\)

Problem 2: \(^{239}_{94}Pu

ightarrow^{4}_{2}He+\)

• Mass - number: \(A = 239 - 4=235\)
• Atomic - number: \(Z=94 - 2=92\)
• The element with \(Z = 92\) is uranium (\(U\)). So the particle is \(^{235}_{92}U\)

Problem 3: \(^{235}_{92}U

ightarrow+\ ^{231}_{90}Th\)

• Mass - number: \(A = 235-231 = 4\)
• Atomic - number: \(Z=92 - 90=2\)
• The particle is \(^{4}_{2}He\)

Problem 4: \(^{1}_{1}H+^{3}_{1}H

ightarrow\)

• Mass - number: \(A=1 + 3=4\)
• Atomic - number: \(Z=1 + 1=2\)
• The particle is \(^{4}_{2}He\)

Problem 5: \(^{6}_{3}Li+^{1}_{0}n

ightarrow^{4}_{2}He+\)

• Mass - number: \(A=6 + 1-4=3\)
• Atomic - number: \(Z=3+0 - 2=1\)
• The element with \(Z = 1\) is hydrogen (\(H\)). So the particle is \(^{3}_{1}H\)

Problem 6: \(^{27}_{13}Al+^{4}_{2}He

ightarrow^{30}_{15}P+\)

• Mass - number: \(A=27 + 4-30=1\)
• Atomic - number: \(Z=13 + 2-15=0\)
• The particle is \(^{1}_{0}n\)

Problem 7: \(^{9}_{4}Be+^{1}_{1}H

ightarrow+^{4}_{2}He\)

• Mass - number: \(A=9 + 1-4=6\)
• Atomic - number: \(Z=4 + 1-2=3\)
• The element with \(Z = 3\) is lithium (\(Li\)). So the particle is \(^{6}_{3}Li\)

Problem 8: \(^{37}_{19}K

ightarrow^{0}_{+1}e+\)

• Mass - number: \(A = 37-0=37\)
• Atomic - number: \(Z=19-1 = 18\)
• The element with \(Z = 18\) is argon (\(Ar\)). So the particle is \(^{37}_{18}Ar\)

Answer:

1. \(^{42}_{20}Ca\)
2. \(^{235}_{92}U\)
3. \(^{4}_{2}He\)
4. \(^{4}_{2}He\)
5. \(^{3}_{1}H\)
6. \(^{1}_{0}n\)
7. \(^{6}_{3}Li\)
8. \(^{37}_{18}Ar\)