Question

Please help due in 4 hours. I only have one shot.

Suppose that placing 0.3 inch of lead in front of a gamma source reduces the count rate from 1125 cps to 554 cps. . What is in g/cm2? The density of lead is 11.4 g/cm3.

Answer 1

I = Io e^-(ux)
Where: I = 554; Io=1125; x = 0.3 inches and rho = 11.4g/cm^3
1 inch = 2.54 cm; therefore, 0.3 inches = 0.3 x 2.54 cm = 0.762 cm; x = 0.762 cm

I/Io = e^-(ux)
Io/I = e^(ux)
ln(Io/I) = ux
u=1/x times ln(Io/I)
u=1/0.762 times ln(1125/554)
u =1.312 x 0.708
u = 0.929
u/rho = 0.6898/11.4 = 0.0815

1/(u/rho) (your (the mass absorption coefficient)^-1 = 53.63 g/cm^2

Answer 2

I = Io e^-(ux)
Where: I = 554; Io=1125; x = 0.3 inches and rho = 11.4g/cm^3
1 inch = 2.54 cm; therefore, 0.3 inches = 0.3 x 2.54 cm = 0.762 cm; x = 0.762 cm

I/Io = e^-(ux)
Io/I = e^(ux)
ln(Io/I) = ux
u=1/x times ln(Io/I)
u=1/0.762 times ln(1125/554)
u =1.312 x 0.7084
u = 0.9294
u/rho = 0.9294/11.4 = 0.0815

1/(u/rho) (your (the mass absorption coefficient)^-1 = 1/0.0815 = 12.27 g/cm^2

Answer 3

my attempt:
I = Io e^-(ux)
Where: I = 554; Io=1125; x = 0.3 inches and rho = 11.4g/cm^3
1 inch = 2.54 cm; therefore, 0.3 inches = 0.3 x 2.54 cm = 0.762 cm; x = 0.762 cm

I/Io = e^-(ux)
Io/I = e^(ux)
ln(Io/I) = ux
u=1/x times ln(Io/I)
u=1/0.762 times ln(1125/554)
u =1.312 x 0.708
u = 0.928
u/rho = 0.928/11.4 = 0.081

1/(u/rho) (your (the mass absorption coefficient)^-1 = 12.3g/cm^2

Answer 4

Answer 5

I = Io e^-(ux)
Where I = 554; Io=1125; x = 0.3 inches and rho = 11.4g/cm^3
1 inch = 2.54 cm; therefore, 0.3 inches = 0.3 x 2.54 cm = 0.762 cm; x = 0.762 cm

I/Io = e^-(ux)
Io/I = e^(ux)
ln(Io/I) = ux
u=1/x times ln(Io/I)
u=1/0.762 times ln(1125/554)
u =1.312 x 0.7083736278
u = 0.9293861
u/rho = 0.9293/11.4 = 0.08152510524

reciprocal of the above quantity will be the answer required...

so the answer is 12.266 g/cm^2