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Alpha Decay Energy Found from Nuclear Masses · Physics Playground

Exercises Relativity, quantum and nuclear

Example from Physics, 22.2 Nuclear Forces and Radioactivity

Alpha Decay Energy Found from Nuclear Masses

Find the energy emitted in the $α$ decay of ²³⁹Pu.

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
Nuclear reaction energy, such as released in $α$ decay, can be found using the equation $E = m c^{2}$ . We must first find $Δ m$ , the difference in mass between the parent nucleus and the products of the decay.

The mass of pertinent particles is as follows

²³⁹Pu: 239.052157 u

²³⁵U: 235.043924 u

⁴He: 4.002602 u.
Solution
The decay equation for ²³⁹Pu is

${}^{239}P u \to {}^{235}U + {}^{4}H e .$
22.41

Determine the amount of mass lost between the parent and daughter nuclei.

$\begin{array}{l} Δ m = m (^{239} P u) - (m (^{239} U) + m (^{4} H e)) \\ Δ m = 239.052157 u - (235.043924 u + 4.002602 u) \\ Δ m = 0.0005631 u \end{array}$
22.42

Now we can find E by entering $Δ m$ into the equation.

$E = (Δ m) c^{2} = (0.005631 u) c^{2}$
22.43

And knowing that $1 u = 931.5 m e V / c^{2}$ , we can find that

$E = (0.005631) (931.5 M e V / c^{2}) (c^{2}) = 5.25 M e V .$
22.44

0 of 2 steps open

How did it go?

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From Physics (1st edition) by OpenStax (Rice University), section 22.2, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.