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When $\beta ^+$ particles are stopped in alluminium they produce $\gamma$ radiation. My question is: is the amount of $\gamma$ radiation (whit alluminium infront of the counter) seen on a geiger muller counter the same as the $\beta^+$ radiation when there is no alluminium infront of the counter. Or in other words can you assume that there is an amount of $\beta^+$ radiation comming from a radioactive source when the geigermuller counter gives a lower value when using 1 mm of alluminium to cover the counter than with no absorbing material infront of the counter (when knowing sure there is no $\alpha$ radiation).

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  • $\begingroup$ Not sure what you mean with "drop". The 1mm aluminum will absorb some of the 511keV photons. $\endgroup$
    – rfl
    Dec 3, 2022 at 15:36
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Dec 3, 2022 at 16:21

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First complication: a Geiger counter is not an efficient detector of gamma rays. Most gamma ray photons pass through undetected.

Second complication: each $\beta^+$ yields either two or three gamma ray photons when it annihilates an electron.

Third complication: gamma ray photons are hard to absorb. They are much more prone to Compton scattering. When they encounter matter, they usually scatter until their energy degrades to the point where photoelectric absorption becomes likely. Thus, in a practical experiment, your Geiger counter will be immersed in a fog of scattered gamma rays of degraded energy.

All this makes relating the rate of gamma detections to the rate $\beta^+$ decays difficult.

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