+1 vote
in Physics by kratos

In a scattering experiment an aluminum foil of thickness 10μm is placed in a beam of intensity 8 × 1012 particles per second. The differential scattering cross-section is known to be of the form

dσ/dΩ = A + B cos2θ

where A, B are constants, θ is scattering angle and Ω is the solid angle. With a detector of area 0.01m2 placed at a distance of 6m from the foil, it is found that the mean counting rate is 50 s−1 when θ is 30◦ and 40s−1 when θ is 60◦. Find the values of A and B. The mass number of aluminum is 27 and its density is 2.7 g/cm2.

1 Answer

+1 vote
by kratos
 
Best answer

Solving the above equations we find

A = 1.57 b/Sr

B = 2.88 b/Sr

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