Answer:
E = 18.54 keV
Explanation:
The given reaction of the tritium beta decay is:
[tex] _{1}^{3} H \rightarrow _{2}^{3} He + e^{-} + v [/tex] (1)
To determine the total energy released in the equation (1) we need first write the energy conservation for that equation:
[tex] m(_{1}^{3} H)c^{2} = m(_{2}^{3} He)c^{2} + m(e^{-})c^{2} + m(v)c{2} [/tex] (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:
[tex] E = [m(_{1}^{3} H) - m(_{2}^{3} He)]c^{2} = [3.0160492 u - 3.0160293 u]c^{2} = 0.0000199 uc^{2} [/tex]
Since 1 uc² = 931.4941 MeV, the energy released in reaction (1) in keV is:
[tex] E = 0.0000199 uc^{2} \cdot \frac{931.4941 MeV}{1 uc^{2}} \cdot \frac{1000 keV}{1 MeV} = 18.54 keV [/tex]
Therefore, the energy released in reaction (1) is 18.54 keV.
I hope it helps you!
