Why Does Aluminum Cause Significant Deflection of Alpha Radiation?

wiss
Messages
1
Reaction score
0
In air, high angle scattering of alpha radiation from Am241 is almost
non existent. When a sample is pulled away from a detector the
detection rate drops as 1/r^2.

But if a thin aluminium foil is inserted in the path almost half of
the radiation disappears somewhere else, it do not reach the detector
even thou the energy spectrum is not below the detection threshold.

In 1 cm of air the nuclei take up approximately 0.1% of the area faced
by the alpha particles, In the case of 0.1 mm aluminium the area would
also be about 0.1% of the area (using a nuclear radius of 10^-14 m).

Why do 50% of the radiation get deflected very much by aluminium? I
would expect 0.1%. Can someone explain this to me?

Jonas
 
Physics news on Phys.org
I am attaching a thumbnail plot of the alpha particle range in air. Alpha particle lose energy by collisions with electrons, not with nuclei. The theory is called the Bethe-Bloch energy loss formula, which applies to just about every charged particle except electrons. The range of particles scales something like grams per cm2 for different materials. In particular, note that a 2 MeV alpha has a range of about 1 cm in air. Because the range is actually proportional to grams per cm2, I have to multiply this by the density of air, 0.0012 grams per cm3, so the range of a 2 MeV alpha is about 0.0012 grams per cm2. Because the density of aluminum is about 2.7 grams per cm3, if I divide 0.0012 g/cm2 by 2.7 g/cm3, I get 0.00044 cm or 4.4 microns for the range in aluminum. How thick was your aluminum foil?
 

Attachments

  • Alpha_Particle_Range.jpg
    Alpha_Particle_Range.jpg
    52.9 KB · Views: 638
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top