How Does Lead Attenuate 0.5 MeV Photons Over 1.3 cm?

It's just a factor that will multiply both intensities.I don't know what else to do with the information given.In summary, lead is a good choice for shielding due to its high density and atomic number, allowing it to effectively attenuate photons by striking atoms. To calculate the fraction of photons absorbed from an attenuated beam of 0.5 MeV photons after passing through 1.3 cm of lead, the equation I = I0e-ut can be used, with the linear attenuation coefficient of lead for 0.5 MeV photons being 0.5 cm-1. The resulting fraction is approximately 0.52 or 3/5, with the intensity of the original beam (I0) remaining
  • #1
tastytau
7
0

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.
 
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  • #2
0.5 MeV is not an intensity. And something went wrong in the exponent.
tastytau said:
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.
 
  • #3
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.
 
  • #4
##e^{-0.5\cdot 1.3} = 0.52## not 3/5. But maybe this fraction just came from too coarse rounding.
tastytau said:
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.
 

1. What is photon attenuation in lead?

Photon attenuation in lead refers to the process by which photons, or particles of light, are absorbed or scattered as they pass through a lead material. This can significantly reduce the intensity of the photon beam.

2. Why is lead commonly used for photon attenuation?

Lead is commonly used for photon attenuation because it is a dense material that is highly effective at absorbing photons. It is also relatively inexpensive and readily available.

3. How does photon energy affect attenuation in lead?

The higher the energy of the photons, the less they will be attenuated by lead. This is because higher energy photons are more likely to pass through the material without being absorbed or scattered.

4. What other factors can affect photon attenuation in lead?

In addition to photon energy, the thickness and density of the lead material can also affect attenuation. Thicker and denser lead will provide greater attenuation compared to thinner and less dense lead.

5. Can other materials be used for photon attenuation besides lead?

Yes, there are other materials that can be used for photon attenuation, such as concrete, steel, and tungsten. However, lead remains a popular choice due to its high density and effectiveness at attenuating photons.

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