The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1*1010 133ba atoms. how many are left after (a) 5 years, (b) 30 years, and (c) 190 years?

a) 7.9x10^9 b) 1.5x10^9 c) 3.9x10^4 To determine what percentage of an isotope remains after a given length of time, you can use the formula p = 2^ (-x) where p = percentage remaining x = number of half lives expired. The number of half lives expired is simply x = t/h where x = number of half lives expired t = time spent h = length of half life. So the overall formula becomes p = 2^ (-t/h) And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are: a) 1.1x10^10 * 2^ (-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9 b) 1.1x10^10 * 2^ (-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9 c) 1.1x10^10 * 2^ (-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4