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Absorption of Nuclear Radiation Essay

It was necessary to determine the operating voltage for the G-M counter. Group 7 put the beta source right underneath the sensor and started with a voltage of 200 V, then slowly added more and more voltage until the rate of electrons counted was constant. (This number was 257 V. To this number, 75 more volts were added for a voltage of 232 V. This was left as a constant for the remainder of the experiment. The background radiation was calculated next. This was done by removing the beta radiation and letting the sensor run for ten minutes to pick up what radiation different sources in the room would let off. The number was was 12.2 counts. This was divided by ten for an average of 12.2 counts/minute. The beta source was replaced , but on the second shelf of the counter. Eight different trials were conducted,
introducing eight different thicknesses of polyethylene absorbers ranging from 0-0.610 g/cm^2, with the 0 g/cm^2 trial not having anything anything in between the beta source and the sensor. Results were recorded.

Experimental data
Table 1: 8 trials of mass thickness
Trial
Mass thickness (g/cm^3)
Amount of Radiation (counts/minute)
Corrected Radiation
(counts/minute)
ln(Corrected Radiation)
1
0
1631±40.386
1618.8±43.878
7.389±0.027
2
0.010
1467±38.301
1454.8±41.794
7.283±0.030
3
0.020
1456±38.158
1443.8±41.650
7.275±0.029
4
0.049
1237±35.171
1224.8±38.664
7.111±0.033
5
0.073
1206±34.728
1193.8±38.220
7.085±0.033
6
0.151
823±28.688
810.8±32.181
6.698±0.040
7
0.305
432±20.785
419.8±24.277
6.040±0.060
8
0.610
75±8.66
62.8±12.153
4.140±0.215

Results

SAMPLE CALCULATIONS

Amount of Radiation-background radiation= Corrected radiation
1631-12.2=1618.8
1206-12.2=1193.8

ln(Corrected radiation) =final answer
ln(1618.8)=7.389
ln(1193.8)=7.085

sqrt(number of counts) = Counting error
sqrt(12.2)=3.493
sqrt(1631)=40.386
sqrt(1206)=34.728

Counted+(-)Counting error-background radiation+(-)counting error-corrected radiation =counted error 1631 +(-)40.386 -12.2 +(-) 3.493 -1618.2 = ±43.878 1206 +(-) 34.728 -12.2 +(-) 3.493 – 1193.8 = ±38.220

ln(Corrected radiation +(-) counted error) – final answer = Final error ln(1618.2 +(-) 43.878) – 7.389 = ±0.027 ln(1193.8 +(-) 38.220) – 7.085 = ±0.033

Tau =-1/slope
τ = -1/-5.180g/cm^2
τ=0.1931cm^2/g

Delta slope/slope = delta tau/tau
Δm/m = Δτ/τ
0.1975/5.180 = Δτ/0.1931
Δτ=0.007362

Discussion and Analysis
The experiment was success. The idea was to see the basic functions of nuclear physics, to understand how absorbers can affect radiation and to practice calculations of errors. The slope of values of the thickness of the absorbers vs. the natural log of the electrons counted gave a very linear graph that helps to learn what the tau of the function. Tau represents the mass thickness attenuation factor and is, in fact, the negative inverse of a graph that is thickness vs. natural log. The generally accepted value of tau is 0.2 cm^2/g. The slope that is given by the graph was -5.180±0.1975g/cm^2, which would relate to tau having a value of 0.1931±0.007362cm^2/g. That’s an error of only 3.57%. The most probable reason for this error would be in human error.

Conclusion
The objectives were met. The group got to know the basic layout of the lab equipment, and the correlation between particles sensed vs. mass thickness was able to be calculated.These practices will help in any future experiment
about nuclear physics and in practicing in the lab generally.


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