Kayley KilcoynePhysics 132Guy RosinMVRadioactivityAbstract: In this experiment we used a Geiger counter to test the ionizing radiation emitted by alpha, beta, and gamma decay. We could then determine the half-life of different elements by using the equations τ =1b and t12=τln(2). We also tested to see which radioactive decay was the strongest by placing the decay in the Geiger counter and then placing different material such as paper, aluminum, and lead over them and then measured the radioactivity. Questions:1.Trial Counts/min1 272 293 324 345 346 347 348 349 4010 28The average of the data is 32.6 and the square root of that is 5.71. By using the equation NB±√32.6. There is only one number that does not fit into this range it is40, but it is only 1.69 away from fitting in this range. All of the other points fall in thisconfidence interval. 2. I think based on my data I think that the gamma decay has the most penetrating power because it was able to go through the aluminum and the led and the Geiger counter was still able to count high levels of radioactivity. I think this decay is the most hazardous to bystanders because they are able to damage your cells and mess o f the replication of DNAin the cell. It is able to turn regular cells into mutant cells and they can become cancerous. 3. The half-life of In-116 is found by using the equation1.000199 to find the B value. Then you put 1/B which finds tau, which is 5025.12. Then to find the half life tau have to multiply that by ln(2) which would give you a half life of 3483.15 seconds.4. To find out how many alpha particles are emitted between t 100 min and 300 min in a sample that has 6.0X10^10 atoms and has a half like of 200 minutes you have to use the equation λ=ln(2)t(12), then when you fill in the variables you get ln(2)/200 =3.46E-3 minutes. This shows the decay constant because you are using the half-life. The you have to do the equation for your t values, 100 min and 300 min. x(100)=6E10(3.46E-3) for the 100 minutes which is 4.245 alpha particles. For the 300 minutes you use the equation X(300)=6E10(3.46E-3) and this gives you 2.124E10 alpha particles. To find the amount emitted between these to times you have to subtract them giving you an answer of 2.121E10 alpha particles. 5. Alpha decay is shown by the equation T This equation shows that during alpha decay there are two protons and two neutrons that are lost by the nucleus of the alpha particle. During Beta decay the nucleus of the atom converts a protoninto a neutron this is shown by the equation . The thirdkind of decay is gamma decay and when this happens the protons and neutrons don’t change in the atom but it does change energy levels because it releases radiation. Gamma decay is shown by the equation .6. Two examples of radioactivity in real life would be Fertilizer and smoke detectors. Fertilizer is radioactive because it contains radioactive elements such as potassium and phosphate and it can also have uranium in it, which is radioactive. Smoke detectors use radioactivity because they emit alpha particles and when smoke particles are released into the air they make the smoke detector ring alarming people of smoke in the air. Conclusion: In this experiment we were able to see how the different decays, alpha beta and gamma acted when we put them under the Geiger counter. We could test how strong the radioactive emissions were based on the different materials we put over them to try and block the emissions. We can conclude that the gamma decay was the strongest because its emissions were able to go through lead and aluminum. They wasn’t very much room for error in this experiment. The only thing that could produce error in this experiment was that the Geiger counter was incorrectly reading the amount of radioactivity, or there was excess radioactivity in the room throwing off the
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