the quetion says "f the half life of Americium-241 is 433 years, &your sample size is 80 grams, how much will remain after 1,732 years?" i don't quite understand how to solve this problem... PLEAS HELP. SOS (gotta test 2morrow, and don't really know how to do these problems THANKS!!
1732 years is exactly 4 half-lives. The amount of Americium-241 will have decreased by half four times. The amount remaining will be (1/2)^4 = 1/16 of the original amount
1/16 of 80 is __?
sorry it took so long for me to reply back, i was giving up hope, and was about to shutdown my computer
To solve this problem, you need to understand the concept of half-life. The half-life of a radioactive substance is the time it takes for half of the initial sample to decay.
In this case, you are given that the half-life of Americium-241 is 433 years. This means that every 433 years, the amount of Americium-241 present will be halved.
To find out how much will remain after 1,732 years, you can use the following steps:
1. Determine the number of half-lives that have passed. Divide the given time (1,732 years) by the half-life of the substance (433 years). In this case, 1,732 years / 433 years = 4.
2. Calculate the fraction of the original sample that remains after each half-life. Since each half-life halves the amount, after one half-life, only 1/2 of the original amount remains. After two half-lives, 1/2 * 1/2 = 1/4 of the original amount remains. After three half-lives, 1/2 * 1/2 * 1/2 = 1/8 of the original amount remains. And so on.
3. Calculate the remaining amount of Americium-241. Multiply the initial sample size (80 grams) by the fraction that remains after the number of half-lives that have passed. In this case, 80 grams * (1/2)^4 = 80 grams * 1/16 = 5 grams.
Therefore, after 1,732 years, approximately 5 grams of Americium-241 will remain from your initial 80-gram sample.