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Number of K electrons that are ejected from lead

  • Thread starter TheBoy
  • Start date
  • #1
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Homework Statement



Determine the number of K electrons ejected from a foil of lead with thickness 0.5 x 10^20 atoms/cm^2. It is bombarded by 10^4 photons that each have an energy of 88keV.

Homework Equations



From a figure in the book, it shows that u/p (7.483 cm^2/g = .7483 m^2/kg). I think this is the mass attenuation coefficient.
Z = 82. I think this is the atomic number and is the number of electrons per atom.

The density of lead is 11360 kg/m^3 and the number of electrons per g is 2.38 x 10^23 electrons/g.

The Attempt at a Solution



We need to find number of electrons.
We know thickness, initial number of photons, their energy, and u/p which is the mass attenuation coefficient I think.

I'm not really sure of an equation that can relate this information to the number of electrons ejected.

I know the answer is 903 K electrons because that's what the book gives.

Can anyone help?
 

Answers and Replies

  • #2
4
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I made a mistake. u/p is .564 m^2/kg for the k electrons
 

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