Verify the Energy Release Given for the Process in Equation

Question

Examlex · Accepted Answer

The Answer is For the process

\mathrm { d } + \mathrm { d } ightarrow { } ^ { 4 } \mathrm { He } + \gamma + 23.8 \mathrm { MeV }

we calculate the energy as

E = \left[ 2 imes m _ { \mathrm { d } } - m _ { \mathrm { a } } ight] c ^ { 2 }

where we have used the mass of the alpha particle rather than the mass of the

{ } ^ { 4 } \mathrm { He }

atom to properly account for electron masses. Substituting appropriate values gives

E = [ 2 imes 2.013553214 \mathrm { u } - 4.00150618 \mathrm { u } ] imes 931.494 \mathrm { MeV } / \mathrm { u } = 23.8 \mathrm { MeV }

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