June 2, 2011

Today we began the day by looking at the previous day's blog. From there, we moved onto pg. 5 where began discussing a half-life and radioactive decay.

A half-life is the amount of time it takes for one-half of a radioactive substance to decay. A half-life can vary in time from a few minutes to millions of years. A majority of radioactive substances decay through radioactive decay which we discussed earlier this weak. Some radioactive decays include: gamma decay, alpha decay, and beta decay. If you wanted to represent this decay in a chemical equation, you do the process that we have covered earlier this weak. For example, in #1, if hydrogen-3 were to decay through beta emission process, it would be represented as:

(3/1)H---->(3/2)He + (0/-1)e

After completing problem 1 and graphing it, we continued on throught the rest of the page. 2A is the same as number one, but 2B is different. Because we know that the half-life is 6, after 6 minutes the percent present is reduced by half, therefore it is 50%. After 12 minutes, the percent present is 25%. This pattern can continue on for ever. Therefore the anser to number 2B is .0976 because after 1 hour of 10, 6 minute half-life's, only .0976% of the substance remains.
For 2C, the equation can be written as N=No*(.5)^n. This can be determined because No represents the mass of the starting substance, while n represents the number of half-life's. By taking .5 to the n'th power, you are accounting for the fact that after each half life you are decreasing the original substance by 50%. Finally, you multiply it by No, in order to find the final mass. Using this equation, we can solve question 3.

3. For the first part, we know that the half-life is 4.5 days long. Since we are given a total of 27 days, you divide 27 by 4.5 to find 6. This is your n value, or the number of half-lifes that took place. The No value is the starting mass which is given to us in the 2nd sentence, 43.8g. From here we can plug in the numbers and solve for N.
-N= 43.8*(.5)^6
-N= .684g
In order to do the second part of question 3, you follow the same steps, except instead o dividing 27/4.5, you do: 33.7/4.5 because you want to find the number of half lifes in 33.7 days not 27 days. The answer for the second part is .243g.

After covering this new material, we proceeded on to a lab, where we were told to find the half-life of a substance by watching it radioactively decay. We performed this by placing cesium-237 into a radioactive detector and recorded its radioactivity every half minute. With this information, we were told to make a scatter plot and determine the substances half-life.

Wednesday, June 1

Today in class we started the topic of Nuclear Bombardment. The main idea that Nuclear transmutations can occur spontaneously or artificially. If it happens artificially there is usually some form of bombardment occurring. There are many examples on page 4. We finished 1-6.

After we finished up with the packet we started on our lab. Our lab was on Radioactivity and Shielding. In it we had see what shields worked best against blocking beta particles and gamma rays. Then once we got that data we had to see which type of radiation did a better job of penetrating through the shields. After we collected all of our data we had an easy reading quiz that consisted of problems one could find on pages 2 and 3 of the packet.