In an experiment done by scattering 5.5 MeV particleson a thin gold foil, studen

Question

In an experiment done by scattering 5.5 MeV particleson a thin gold foil, students find that 10,000 particles arescattered at an angle greater than 50 degrees. How many particles will be scattered between 70 degreesand 80 degrees? The answer is 1347, but I can't seem to figure it out. A formula that might help is: df =-nt[(Z1Z2e2)/(80K)]2cot(/2)csc2(/2)d In an experiment done by scattering 5.5 MeV particleson a thin gold foil, students find that 10,000 particles arescattered at an angle greater than 50 degrees. How many particles will be scattered between 70 degreesand 80 degrees? The answer is 1347, but I can't seem to figure it out. A formula that might help is: df =-nt[(Z1Z2e2)/(80K)]2cot(/2)csc2(/2)d

Explanation / Answer

f() = number scattered at angle > = Ab^2(A=constant, b=impact parameter) b is related to scattering angle by b = B cot(/2) >>> B is another constant. so f() = C [cot(/2)]^2 >>> C is a constantrelated to A and B we know f(50) = 10000 = C [cot(25)]^2 C = 10000/[cot(25)]^2 # scattered between 70 and 80 degrees = f(70) - f(80) = C [ cot(35)^2 - cot(40)^2] = (10000/[cot(25)]^2)[ cot(35)^2 - cot(40)^2] = 1347 =

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