# Absorption of Nuclear Radiation Essay

## Absorption of Nuclear Radiation

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.