Question

What fraction of 5 MeV α particles will be scattered through angles greater than 2.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thickness 10-8 m?

Answer #1

n = pN_{a}/M

Where N_{a} = 6.022*10^{23} is avogadro's no.

p is density of gold = 19.3 g/cm^{3} = 19.3*10^{6}
g/m^{3}

M is molecular mass of gold = 196.97 g/mole

n = [(19.3*10^{6} )*( 6.022*10^{23})/(196.97)] =
5.9*10^{28} atoms/m^{3}

Now fraction of
particles scattered by angle

Where E is energy of
particle = 5*10^{6}*(1.6*10^{-19}) =
8*10^{-13} J

z_{1} is atomic no. of gole = 79

Z_{2} is atomic no. of
particle = 2

t is thickness of sheet = 10^{-8} m

f =
(1.85*10^{21})*(5.167*10^{-28})*(2100.329)

f = 2.008*10^{-3}

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Female
63
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5
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Female
61
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