Question

for questions 8 - 12, balance the nuclear equation, then indicate if the equation represents a fission or a fusion. 8. 231 91pa→192 77ir+ 9. 4 2he+1 2h→ 10. 9 4be+12 6c→ 11. 238 92u+1 0n→128 52te+ 12. 262 105db→96 42mo+28 14si+

Explanation:

Step1: Recall nuclear - equation rules

In a nuclear equation, the sum of mass numbers and atomic numbers on both sides must be equal.

Step2: Balance equation 8

For \(_{91}^{231}Pa
ightarrow_{77}^{192}Ir + X\).
Mass - number of \(X\): \(231-192 = 39\).
Atomic - number of \(X\): \(91 - 77=14\). So \(X = _{14}^{39}Si\). This is a fission reaction as a heavy nucleus (\(Pa\)) splits into lighter nuclei (\(Ir\) and \(Si\)).

Step3: Balance equation 9

For \(_{2}^{4}He+_{1}^{2}H
ightarrow X\).
Mass - number of \(X\): \(4 + 2=6\).
Atomic - number of \(X\): \(2+1 = 3\). So \(X=_{3}^{6}Li\). This is a fusion reaction as lighter nuclei combine to form a heavier nucleus.

Step4: Balance equation 10

For \(_{4}^{9}Be+_{6}^{12}C
ightarrow X\).
Mass - number of \(X\): \(9 + 12=21\).
Atomic - number of \(X\): \(4+6 = 10\). So \(X = _{10}^{21}Ne\). This is a fusion reaction.

Step5: Balance equation 11

For \(_{92}^{238}U+_{0}^{1}n
ightarrow_{52}^{128}Te+X\).
Mass - number of \(X\): \(238 + 1-128=111\).
Atomic - number of \(X\): \(92+0 - 52 = 40\). So \(X=_{40}^{111}Zr\). This is a fission reaction.

Step6: Balance equation 12

For \(_{105}^{262}Db
ightarrow_{42}^{96}Mo+_{14}^{28}Si+X\).
Mass - number of \(X\): \(262-(96 + 28)=138\).
Atomic - number of \(X\): \(105-(42 + 14)=49\). So \(X=_{49}^{138}In\). This is a fission reaction.

Answer:

1. \(_{91}^{231}Pa

ightarrow_{77}^{192}Ir+_{14}^{39}Si\), Fission

1. \(_{2}^{4}He+_{1}^{2}H

ightarrow_{3}^{6}Li\), Fusion

1. \(_{4}^{9}Be+_{6}^{12}C

ightarrow_{10}^{21}Ne\), Fusion

1. \(_{92}^{238}U+_{0}^{1}n

ightarrow_{52}^{128}Te+_{40}^{111}Zr\), Fission

1. \(_{105}^{262}Db

ightarrow_{42}^{96}Mo+_{14}^{28}Si+_{49}^{138}In\), Fission