Question

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Answer 1

Answer on Question #42889, Physics, Nuclear Physics

10. For the following fission reaction

$${ } ^ { 2 3 5 } \mathrm { U } + \mathrm { n } \rightarrow { } ^ { 1 4 0 } \mathrm { C e } + { } ^ { 9 4 } \mathrm { Z r } + 2 \mathrm { n } ,$$

Find the disintegration energy.

Solution:

Given:

$$M _ {u} = 2 3 5. 0 2 \mathrm {u},$$ $$M _ {n} = 1. 0 \mathrm {u},$$ $$M _ {c e} = 1 3 9. 9 \mathrm {u},$$ $$M _ {z r} = 9 3. 9 \mathrm {u},$$ $$Q = ?$$

The disintegration energy $Q$ is the energy transferred from mass energy to kinetic energy of the decay products.

$$Q = m _ {i} c ^ {2} - m _ {f} c ^ {2}$$

Initial mass is

$$m _ {i} = M _ {u} + M _ {n}$$

The final mass is

$$m _ {f} = M _ {c e} + M _ {z r} + 2 M _ {n}$$

Thus,

$$Q = \left(M _ {u} - M _ {c e} - M _ {z r} - M _ {n}\right) c ^ {2}$$ $$Q = (2 3 5. 0 2 - 1 3 9. 9 - 9 3. 9 - 1. 0) \cdot 9 3 1. 5 \mathrm {M e V / u} = 2 0 4. 9 3 \mathrm {M e V} \approx 2 0 5 \mathrm {M e V}$$

Answer: $Q = 205 \mathrm{MeV}$

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