# Beta and Gamma (Physics 377)

QUESTIONS

L Comment on the semi log graph of the data from data table 1 for intensity as function of  thickness of lead absorber. At what thickness are all the $\beta$ rays absorbed? After the $\beta$ rays are absorbed does the graph of absorption of $\gamma$ rays only show a linear behavior?

Lead is a very heavy metal (big atomic number). Because of this it absorbs very fast the β rays of low energy which are electrons emitted from the nucleus. Here the β rays are absorbed in the first layers from the surface. We see that the first part of the graph is more sloped; this is the region where absorption of beta rays takes place. From the graph its thickness is d=1700-1800 mg/cm^2 which for lead is about d=0.05 inch

After the β rays from the beam are absorbed, it remains in the beam just the γ rays which are photons of high energy. They are absorbed accordingly to the equation I(x)=I_0 e^(-μ*x) where μ=n*σ is the absorption coefficient (proportional to the concentration n and the nuclear absorption cross-section σ). Therefore the graph of log N(counts) vs x (distance) should be linear and have another slope from its first portion for γ rays (after all β rays have been absorbed).

2. Comment on the semi log graph of the data from Data Table 2 of the intensity of $^{60}Co$ radiation as a function of thickness of polyethylene absorber. Does the absorption of the \$beta$ rays take place over several absorbers? At what thickness are all of the $\beta$ rays absorbed? After the $\beta$ rays are absorbed, does the graph of absorption of $\gamma$ rays only show a linear behavior?

Also in this case, having the same $^{60} Co$ initial beam of radiation, but a lighter absorber (polyethylene) the β rays are again absorbed in the first layers of the material. The difference is that the absorption coefficient for β(and also for γ) rays is smaller (lighter atom, smaller cross-section σ) and therefore the thickness of the absorbing layer is bigger. Again like in the first situation when in the beam remain just the γ rays the graph should become linear with the difference that the rate of absorption is smaller.

The graph is full of experimental errors and shows no linearity but fitting with two lines of different slopes the data it seems that the first slope extends up until a thickness of d=600 mg/cm^2 which for polyethylene means a about d=0.1 inch . Indeed this is a bigger length of absorption for beta rays than using lead.

3. Comment on the data in Data Table 3 for the intensity of $^{90}Sr$ radiation versus the thickness of lead absorber. What is your conclusion about the absorption of $\beta$-rays in lead?

$^90 Sr$ emits radiations that is a bit higher in energy than that emitter by $^{60} Co$. The similarity between the two emitted radiations is also that they are the same composed of β and γ rays. Like in all other situations β rays are absorbed in the first layers of the material (have lowest energy).  The graph should be composed of thw lines having different slopes. After the initial distance where beta rays are absorbed the graph of log N(counts) vs x changes the slope. The slope should be about the same for the two lines as in Q1). The explanation is simple: the absorbing material is the same: lead.

From data we see indeed that beta rays are absorbed into lead in the first layers of the material: first slope extends up until is $d=1700-1800 mg/cm^2$ which for lead is about $d=0.05 inch$. This thickness is equal to the one found in Q1)

4. Comment on the semi log graph of the data from Data Table 4 for the intensity of $^{90} Sr$ radiation as function of thickness of polyethylene absorber. Is the graph approximately linear? If it is not linear over the whole range, is it at least linear over some portion of the range?

The same theoretical considerations as for Q2 apply with the only difference that now the intensity of the initial radiation is somewhat bigger (Sr source). The absorbing material is polyethylene and theoretically the beta rays should be absorbed in a thicker layer at the surface than for lead and the  absorption coefficient for gamma rays should be smaller than using lead. The graph is clean and shows indeed two lines of different slope. The first slope extends up until a thickness of d=600 mg/cm^2 which for polyethylene means a about =0.1 inch . The thickness of beta absorbing layer is about the same as in Q3 (Co source with polyethylene)

5. Compare the slope of the graphs, for $^{60}Co$ in the two absorbers. Using that as a basis for comparison, estimate how much more strongly the lead absorbs $\gamma$-rays than does the polyethylene.

As it was explained above, the slope of the linear part of the graph where the gamma rays are absorbed is equal to the absorption coefficient. For polyethylene (lighter atoms) the absorbing cross-section is smaller than for lead. It means that polyethylene has a smaller absorption coefficient, and therefore the corresponding graph slope is smaller (the graph is more inclined for lead).

The slopes of the two lines indicating beta absorption are the same for Co And Sr sources. Also the slopes of the two lines indicating gamma absorption are the same (but smaller than for beta absorption) for Co and Sr sources.

For polyethylene the slope for gamma absorption is about

$slope=(7-3)/(900-400)=4/500=8*10^{-3}$   $[ln⁡(N)/(d (mg/cm^2 ))]$

For lead the slope for gamma absorption is about

$slope=(5.5-5)/(8000-1800)=0.5/6200=8*10^{-5}$   $[ln⁡(N)/(d (mg/cm^2 ))]$
This means lead is a 100 times a better shielding material.

6. For $\gamma$ rays of the approximate energy of the $^{60}Co$ $\gamma$ rays in lead, the approximate value of the absorption coefficient is $\mu = 0.65 cm^{-1}$ and with $\rho_{p b} = 11.4 gm/cm^3$ , then $\mu_m = 5.7 x 10^{-2} cm^2/g$. Consider this as the accepted value, what is the accuracy of your measurement of $\mu_m$.

The equation to be used here is

$μ_m=μ/ρ_Pb =0.65/11.4=5.7*10^{-2}$   $(cm^2)/g$

From the slope of the linear we obtain a value of about $μ_m=8*10^{-5}*10^3=8*10^{-2}$  $(cm^2)/g$ which is a fairly good value.