Kinetic energy of an electron on the β spectrum?

AI Thread Summary
The discussion revolves around calculating the maximum kinetic energy of an electron emitted during the β- decay of a 191Os nucleus to an excited state of 191Ir. The Q value of the reaction was determined to be 0.3139 MeV, and after subtracting the energy of the excited state (171 keV), the maximum kinetic energy calculated is 0.143 MeV. There is uncertainty about the correctness of this answer, but it aligns with the understanding that the maximum kinetic energy equals the Q value minus the energy of the excited state. The calculations and reasoning presented support the conclusion that 0.143 MeV is the correct answer. The discussion emphasizes the importance of accurately determining the Q value in such decay processes.
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Homework Statement



A 191Os nucleus decays via β- emission to an excited state of 191Ir at 171 keV. What is the maximum kinetic energy of the electron on the β spectrum?
a) 0.134 MeV
b) 0.143 MeV
c) 0.171 MeV
d) 1.859 MeV

Homework Equations



191 Os = 190.960928u
191 Ir = 190.960591u

(190.960928u - 190.960591u) * 931.5MeV/u = 0.3139 MeV

0.3139 MeV - 0.171 MeV = 0.143 MeV

The Attempt at a Solution


[/B]
I am unsure if my answer of 0.143 MeV is correct. In my notes it says the maximum kinetic energy is equal to the Q value of the reaction, so I wokred out the Q value above then subtracted the "excited state" value.
 
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