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PHY3708 Assignment 3 Solutions 2026

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PHY3708 Assignment 3 Solutions 2026 0-7-9-3-2-2-6-4-2-7 unisa

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PHY3708 Assignment 3 Solutions 2026
UNISA

, Question 1: Prompt energy released during the fission process

The assignment gives the fission reaction:
1 235 140 ∗ 96 ∗ 139 95 1
0𝑛 + 92 𝑈 → 54 𝑋𝑒 + 38 𝑆𝑟 → 54 𝑋𝑒 + 38 𝑆𝑟 +2 0𝑛 + 7𝛾


The two prompt neutrons have a total kinetic energy of 5.2 MeV, while the seven
prompt gamma rays have a total energy of 6.7 MeV.

Calculate the Q-value

The energy released is obtained from the mass difference:

𝑄 = (𝑀initial − 𝑀final )𝑐 2


Using standard atomic masses:
235
• 𝑚( 𝑈) ≈ 235.04393 𝑢

• 𝑚(𝑛) ≈ 1.008665 𝑢
139
• 𝑚( 𝑋𝑒) ≈ 138.92133 𝑢
95
• 𝑚( 𝑆𝑟) ≈ 94.92172 𝑢

Therefore,

𝑀initial = 235.04393 + 1.008665
𝑀initial = 236.052595 𝑢


The final mass is

𝑀final = 138.92133 + 94.92172 + 2(1.008665)
𝑀final = 235.860385 𝑢


Thus,

Δ𝑀 = 236.052595 − 235.860385
Δ𝑀 = 0.192210 𝑢


Using

1 𝑢 = 931.494 MeV/𝑐 2


we obtain

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