How Does Lead Attenuate 0.5 MeV Photons Over 1.3 cm?

[tastytau]

Homework Statement

Calculate the fraction of photons absorbed from an attenuated beam of 0.5 MeV photons after it has gone through 1.3 cm of lead? The linear attenuation coefficient of lead for 0.5 MeV photons is 0.5 cm-1. Why is lead a good choice as a material for shielding?

Homework Equations

I = I₀e^-ut

The Attempt at a Solution

[/B]
I = 0.5 MeV e^{-(0.5/cmx1.3cm)}

I = 0.26 = 0.3 MeV

Fraction = 3/5

Lead is effective at attenuating photons because it has a high density and high atomic number. Many photons are attenuated in lead by striking atoms.

 

[mfb]

0.5 MeV is not an intensity. And something went wrong in the exponent.

Lead is effective at attenuating photons because it has a high density and high atomic number. Many photons are attenuated in lead by striking atoms.

Good.

 

[tastytau]

What was wrong in the exponent? I have -(0.5 cm^-1 x 1.3 cm)

I don't know what else to do with the information given.

 

[mfb]

$e^{-0.5\cdot 1.3} = 0.52$ not 3/5. But maybe this fraction just came from too coarse rounding.

I don't know what else to do with the information given.

Just keep I₀ as unknown intensity. You are calculating a ratio of two intensities anyway, it will cancel in this ratio.