This is in a nuclear radiation lab.

(gamma rays from Cs 137 passing through lead.

initial radiation (with one plate) 500cpm

final radiation intensity with 2 plates 300cpm

thickness of 1 absorbing plate in centimeters 0.635

I need to figure out the penetration factor and i need to figure the thickness required to block 99 percent of the gamma rays. I'm getting .40 and 1.257cm respectively. Please tell me what it actually should be and how to figure it out, i'm pretty certain it is wrong.

To calculate the penetration factor and the thickness required to block 99 percent of the gamma rays, we can use the formula:

Penetration Factor (PF) = (Final Radiation Intensity) / (Initial Radiation Intensity)

In this case, the initial radiation intensity is 500 cpm, and the final radiation intensity with 2 plates is 300 cpm.

PF = 300 cpm / 500 cpm = 0.6

The penetration factor is 0.6.

Now, to determine the thickness required to block 99 percent of the gamma rays, we can use the equation:

Transmitted Intensity = Initial Intensity * e^(-μx)

Where:
- Transmitted Intensity is the intensity of the radiation passing through the material.
- Initial Intensity is the initial radiation intensity.
- μ is the linear absorption coefficient, which is specific to the material and the radiation type.
- x is the thickness of the material.

Since we want to block 99 percent of the gamma rays, the transmitted intensity will be 1 percent of the initial intensity:

Transmitted Intensity = 0.01 * Initial Intensity

Now, we rearrange the equation to solve for x:

x = -ln(Transmitted Intensity / Initial Intensity) / μ

Given the values provided, let's proceed with the calculations.

First, we need to find the linear absorption coefficient (μ) for lead and gamma rays from Cs 137. The linear absorption coefficient depends on the energy of the gamma rays.

The energy of gamma rays from Cs 137 is approximately 0.662 MeV. For this energy, the linear absorption coefficient for lead is approximately 0.015 cm^-1.

Now, we can calculate the thickness required to block 99 percent of the gamma rays:

x = -ln(0.01) / 0.015 cm^-1

x ≈ 4.6 cm

Therefore, the thickness required to block 99 percent of the gamma rays is approximately 4.6 cm (not 1.257 cm).

To summarize:
- The penetration factor (PF) is approximately 0.6.
- The thickness required to block 99 percent of the gamma rays is approximately 4.6 cm.