Meet students taking the same courses as you are!Join a Numerade study group on Discord



Numerade Educator



Problem 43 Hard Difficulty

The isotope $\stackrel{198}{79} \mathrm{Au}$ (atomic mass 197.968 u) of gold
has a half-life of 2.69 days and is used in cancer therapy. What
mass (in grams) of this isotope is required to produce an activity of 315 Ci?


$1.29 \times 10^{-3} \mathrm{g}$


You must be signed in to discuss.

Video Transcript

so the activity has a form of Lambda Times, and now we are given that the activity is 315 Curie. Um, and then we can find 1/2 life from the from the rate of decay. So a is 315 curie, and then one Curie is equal to you. Take a look at the book. It's 3.7, uh, times 10 to the 10th back. Carell's So that's one Curie. This is able to remember the decay constant easy to Ellen of two divided by the half life. So we're given 1/2 life in days, so we need to convert to 0.69 days in two seconds. Um, so there are 24 hours per day and 3600 seconds per hour. Okay. And all of that is dividing Ellen of two. And then that gives us in somewhat supplying everything out. We get an end of 3.9, um, one times, 10 to the 18th. Um, Adams. So now how to find the weight in grams, so we're going to convert. So they're 3.91 times. Well, first off, Okay, so we just need to multiply this number uh, 3.91 times 10 to the 18 times the mass of each Adam, which is equal to 197.968 You remember one. You leave the 1.66 year old five times, 10 to the negative. 27 kilograms. Um, that's for one. You and there are 1000 grounds prayer, one kilogram multiplying all of this out. We get a total mass of 1.29 times 10 to the negative third grounds.