can you help me with this .
20 Iodine – 125 emits gamma rays when it decays to Tellerium-125 and has a half-life of 60 days. If a 0.020 g pellet of iodine – 125 is implanted into a prostate gland, how much iodine – 125 remains there after one year?
21. Barium – 122 has a half-life of 2 minutes. Suppose you obtain a sample weighing 10.0 g and it takes 10 minutes to set up an experiment. How many grams of Barium – 122 will remain at the point when you begin the experiment?
22. A normal adult jawbone contains 200 mg of Carbon-14 in a living person. If scientists found a jawbone that only had 50mg of Carbon-14, how old is the bone? (The half-life of C-14 is 5730 years).
For question 20:
After 1 year (365 days), there will be 365/60 = 6 half-lives that have passed.
Starting with 0.020 g of iodine-125:
After 1 half-life: 0.020 g / 2 = 0.010 g
After 2 half-lives: 0.010 g / 2 = 0.005 g
After 3 half-lives: 0.005 g / 2 = 0.0025 g
After 4 half-lives: 0.0025 g / 2 = 0.00125 g
After 5 half-lives: 0.00125 g / 2 = 0.000625 g
After 6 half-lives: 0.000625 g / 2 = 0.0003125 g
Therefore, after 1 year, there will be 0.0003125 g of iodine-125 remaining in the prostate gland.
For question 21:
After 10 minutes, 5 half-lives of Barium-122 have passed (10 minutes / 2 minutes = 5 half-lives).
Starting with 10.0 g of Barium-122:
After 1 half-life: 10.0 g / 2 = 5.0 g
After 2 half-lives: 5.0 g / 2 = 2.5 g
After 3 half-lives: 2.5 g / 2 = 1.25 g
After 4 half-lives: 1.25 g / 2 = 0.625 g
After 5 half-lives: 0.625 g / 2 = 0.3125 g
Therefore, when you begin the experiment after 10 minutes, there will be 0.3125 g of Barium-122 remaining.
For question 22:
To find the age of the bone, we need to determine the number of half-lives that have passed from 200 mg to 50 mg of Carbon-14.
1 half-life: 200 mg / 2 = 100 mg
2 half-lives: 100 mg / 2 = 50 mg
Therefore, 2 half-lives have passed, which corresponds to 2 x 5730 = 11460 years. Thus, the bone is approximately 11460 years old.