Nuclear radiation emitted from a Co 60 radioactive source is detected by a photomultiplier tube (PMT) coupled to a scintillator crystal. Which of the following option(s) is/are correct?
γ radiation from Co 60 will directly hit the photocathode of the PMT without interacting with the scintillator crystal and produce a signal
β radiation from Co 60 source interacts with the scintillator crystal, producing γ radiation, which will hit the photocathode of the PMT and produce a signal
A mu-metal shield is put all around the PMT to nullify the effect of external electric fields
A mu-metal shield is put all around the PMT to nullify the effect of external magnetic fields
Correct answer: (D) A mu-metal shield is put all around the PMT to nullify the effect of external magnetic fields
Explanation
The scintillator converts deposited ionizing-radiation energy into optical photons detected by the PMT. Mu-metal has high magnetic permeability and shields magnetic fields, rather than electric fields.