An 3h nucleus beta decays into 3he by creating an electron and an antineutrino according to the following reaction.
1^3(> 2^3(he)+e-+v
use appendix b to determine the total energy released in this reaction.

E = 18.54 keV

Explanation:

The given reaction of the tritium beta decay is:

  (1)

To determine the total energy released in the equation (1) we need first write the energy conservation for that equation:

  (2)

From equation (2):

m(v) = 0 and,

m(e⁻) << m(³H) and m(³He), so its mass can be neglected  

The energy released in the reaction (1) is:

Since 1 uc² = 931.4941 MeV, the energy released in reaction (1) in keV is:       

Therefore, the energy released in reaction (1) is 18.54 keV.

I hope it helps you!

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