# Q-Value Electron Capture Kr-81

1. Jun 1, 2015

### says

1. The problem statement, all variables and given/known data
Calculate the Q-value for the electron capture beta decay of Kr-81
(Answer in MeV, correct to 6 significant figures)

Atomic Masses (amu)
Kr-81 = 80.916592(3)
Br-81 = 80.916291(3)

proton 1.00727647
neutron 1.00866501
electron 0.0005485803
2. Relevant equations
Q = ∑mic2-∑mfc2

Electron Capture = ZX +e_ ---> Y+ neutrino

3. The attempt at a solution
Q = (80.916592 + 0.0005485803) - 80.916291
= 0.00084958029 * 931.5020
= 0.7913857 MeV

I found a Q value of .280801 Mev online. Not sure why I'm off by so much...

Q = 80.916592 - 80.916291 = 0.00030099999

Q = 0.00030099999 * 931.5020cMeV
= 0.28038209268
= 0.2803821 Mev (correct to 6 sig figs)

Seems better, but I've calculated this using the atomic masses and not the nuclear masses...

Last edited: Jun 1, 2015
2. Jun 1, 2015

### BvU

one me off. Sets you thinking !
The electron is already in the atomic mass, so you don't add it: that's double counting !
See here (again !)

3. Jun 1, 2015

### says

I've calculated this using the atomic masses and not the nuclear masses though. The answer below looks correct. Do I not have to worry about getting rid of electrons when calculating the Q-value in K-capture?

Q = 80.916592 - 80.916291 = 0.00030099999

Q = 0.00030099999 * 931.5020cMeV
= 0.28038209268
= 0.2803821 Mev (correct to 6 sig figs)

4. Jun 1, 2015

### says

Ok, I tried to calculate Q with the nuclear masses ( atomic mass - mass of electrons)

Q = (( 80.916592 - 0.0192003105* ) - ( 80.916291-0.0192003105 )) * 931.5020

* Initial mass = atomic mass - mass of electrons + mass of one electron that moves into the nucleus.

Q = 0.00030099999 * 931.5020
= 0.28038210199 MeV

5. Jun 2, 2015

### BvU

Yes.
Q = 80.916592 - 80.916291 = 0.00030099999 ? No it's not: it's 0.000301 and now you only have 3 digit accuracy, 280 MeV oops, keV !
(280.8 +- 0.5 here on p. 1624) No way you can find that with 6 digit accuracy anywhere (but again, that's the exercise composers' problem).

Last edited: Jun 2, 2015
6. Jun 2, 2015

### BvU

I suppose that, since the composer of the exercise specified this accuracy it is wisest to stay with the 280.383 keV that you found, so that it is clear from your answer that you did the correct counting. As long as you realize that in fact there are only 3 significant digits.

Is it all a bit clear to you now ?
Mind you, explicitly putting nuclear mass = atomic mass - Z * electrons mass is not right because you ignore the binding energies of all the electrons. That you do it on both sides is what saves the numerical value of answer, but ignoring all of them instead of only the one that is involved in the process is too coarse to claim $m_N = m - Zm_e$.