Question

Using the table of linear attenuation coefficients provided below, calculate the thickness of iron and lead required to attenuate narrow, collimated, monochromatic beams of 0.1 Mev and 1.0 Mev gamma rays to:
(a) One-half the incident intensity (HVL) 
(b) One-tenth the incident intensity (TVL)
(c) What is the mathematical relationship between a HVL and a TVL?

	Linear Attenuation Coefficient (μL), cm-1
	0.1 MeV	1.0 MeV
Iron	2.72	0.470
Lead	59.7	0.771

Answer 1

a]

The intensity of the beam after passing through the material is:

=>

For Lead when Energy = 0.1 MeV

when Intensity is halved, I/I_o = 1/2

taking ln on both sides gives

=> x = 0.2548 cm

Repeat for 1 MeV for Iron and Lead

b]

when Intensity is one-tenth, I/I_o = 1/10

therefore,

=> x = 0.8465 cm

repeat the same for E = 1 MeV for Iron and Lead

c]

for one-half:

for one-tenth,

but ln(10) = ln(2 x 5) = ln 2 + ln 5

therefore,

=>

this is the relation between HVL and TVL.