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Problem 33 Easy Difficulty

A $^{3} \mathrm{H}$ (tritium) nucleus beta decays into $^{3} \mathrm{He}$ by creating an electron and an antineutrino according to the reaction
$$_{1}^{3} \mathrm{H} \quad \rightarrow \quad_{2}^{3} \mathrm{He}+\mathrm{e}^{-}+\overline{\nu}$$ Use Appendix B to determine the total energy released in this reaction.

Answer

0.0186 \mathrm{MeV}

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Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg

Video Transcript

the number 33 were asked how much energy is released in this reaction. So I just look up the mass of these in the back in appendix B. For this, I get 3.16 049 Uh, for the daughter. Here I get three 0.16 0 to 9. And these don't have mass. That's just a single electron. And this is a neutrino. Oh, so I subtract and I get that the missing amount of mass is print. 0000 too. And these were all measured in use atomic mass units. So now I'm just going to convert that I'm going to convert that to the energy equivalent one. You is the same as a 9 31.5 mega electron volts. And I get that. The energy was point 0186 Mega electron volts

University of Virginia
Top Physics 103 Educators
Christina K.

Rutgers, The State University of New Jersey

LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Jared E.

University of Winnipeg