Example 32.3 from College Physics, 32.6 Fission

Calculating Energy Released by Fission

Calculate the energy released in the following spontaneous fission reaction:

^{238} U \to^{95} Sr +^{140} Xe + 3 n

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given the atomic masses to be $m (^{238} U) = 238.050784 u$ , $m (^{95} Sr) = 94.919388 u$ , $m (^{140} Xe) = 139.921610 u$ , and $m (n) = 1.008665 u$ .

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
As always, the energy released is equal to the mass destroyed times $c^{2}$ , so we must find the difference in mass between the parent $^{238} U$ and the fission products.
Solution
The products have a total mass of

$\begin{array}{l} m_{products} & = & 94.919388 u + 139.921610 u + 3 (1.008665 u) \\ = & 237.866993 u. \end{array}$
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The mass lost is the mass of $^{238} U$ minus $m_{products}$ , or

$Δ m = 238.050784 u - 237.8669933 u = 0.183791 u,$
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so the energy released is

$\begin{array}{l} E & = & (Δ m) c^{2} \\ = & (0.183791 u) \frac{931.5 Me V/ c^{2}}{u} c^{2} = 171.2 MeV. \end{array}$
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Discussion
A number of important things arise in this example. The 171-MeV energy released is large, but a little less than the earlier estimated 240 MeV. This is because this fission reaction produces neutrons and does not split the nucleus into two equal parts. Fission of a given nuclide, such as $^{238} U$ , does not always produce the same products. Fission is a statistical process in which an entire range of products are produced with various probabilities. Most fission produces neutrons, although the number varies with each fission. This is an extremely important aspect of fission, because neutrons can induce more fission, enabling self-sustaining chain reactions.

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