Question

An 3H nucleus beta decays into 3He by creating an electron and an antineutrino according to the following reaction.

1^3(H)--> 2^3(He)+e-+v
Use Appendix B to determine the total energy released in this reaction.
_____keV

Answer 1

E = 18.54 keV

Step-by-step explanation:

The given reaction of the tritium beta decay is:

`_(1)^(3) H \rightarrow _(2)^(3) He + e^(-) + v` (1)

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

`m(_(1)^(3) H)c^(2) = m(_(2)^(3) He)c^(2) + m(e^(-))c^(2) + m(v)c{2}` (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:

`E = [m(_(1)^(3) H) - m(_(2)^(3) He)]c^(2) = [3.0160492 u - 3.0160293 u]c^(2) = 0.0000199 uc^(2)`

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

`E = 0.0000199 uc^(2) \cdot (931.4941 MeV)/(1 uc^(2)) \cdot (1000 keV)/(1 MeV) = 18.54 keV`

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

I hope it helps you!