- #1
Federica
- 12
- 1
Hi all,
I have some doubts regarding the experiment of Madame Wu. I know a strong magnetic field is used to polarise the ##^{60}Co## nuclei, then we have an anthracene scintillator on the top of the sample which will detect the electron produced in the decay: ##^{60}Co \rightarrow ^{60}Ni^{**} + e^{-} + \bar{\nu}_{e}##. Two others scintillators will detect the two gammas produced by the de-excitation of ##^{60}Ni^{**}##, one is set at 90° to the field (the 'equatorial counter') and the other one at 0° ('the polar counter').
Now, since I suppose the spins are parallel to the field, I would expect the polar counter to detect more photons than the equatorial one (as long as we have polarisation), but the opposite happens. And since the anti-neutrino is right-handed, the electron should be left-handed, which means the anthracene detector should detect more electrons when the field is on the downside. But once again, the opposite happens.
Can someone help me to understand what's wrong with my reasoning?
I have some doubts regarding the experiment of Madame Wu. I know a strong magnetic field is used to polarise the ##^{60}Co## nuclei, then we have an anthracene scintillator on the top of the sample which will detect the electron produced in the decay: ##^{60}Co \rightarrow ^{60}Ni^{**} + e^{-} + \bar{\nu}_{e}##. Two others scintillators will detect the two gammas produced by the de-excitation of ##^{60}Ni^{**}##, one is set at 90° to the field (the 'equatorial counter') and the other one at 0° ('the polar counter').
Now, since I suppose the spins are parallel to the field, I would expect the polar counter to detect more photons than the equatorial one (as long as we have polarisation), but the opposite happens. And since the anti-neutrino is right-handed, the electron should be left-handed, which means the anthracene detector should detect more electrons when the field is on the downside. But once again, the opposite happens.
Can someone help me to understand what's wrong with my reasoning?