Asked by Mrs M
Barium – 122 has a half-life of 2 minutes. Suppose you obtain a sample weighing 100.0 g and it takes 14 minutes to set up an experiment. How many grams of Barium – 122 will remain at the point when you begin the experiment?
Answers
Answered by
DrBob222
NOPE. Perhaps the word approximate gives pause but that isn't even close.
If Ba-122 has a half life of 2 minutes then in 14 minutes it will go through 7 half lifes so if you start with 100 grams it will decay as follows:
START with 100.0 grams
At the beginning you will have 100.0 g and zero minutes has elapsed.
After 1 half life you will have 50 g left and 2 minutes has elapsed.
After 2 half life you will have 25 g left and 4 minutes has elapsed.
After 3 half lives you will have 12.5 g left and 6 minutes has elapsed.
After 4 half lives you will have 6.25 g left and 8 minutes has elapsd.
After 5 half lives you will have 3.125 g left and 10 minutes has elapsed.
After 6 half lives you will have 1.5625 g left and 12 minuies has elapsed.
After 7 half lives you will have 0.78125 g left and 14 minutes has elapsed.
Or you can do it mathematically as follows:
k = 0.69315/2 min = 0.346575
ln(No/N) = kt
ln (100/N) = 0.346575*14 = 4.85205
100/N = 128
N = 100/128 = 0.78125 g left after 14 minutes and just as the experiment starts.
If Ba-122 has a half life of 2 minutes then in 14 minutes it will go through 7 half lifes so if you start with 100 grams it will decay as follows:
START with 100.0 grams
At the beginning you will have 100.0 g and zero minutes has elapsed.
After 1 half life you will have 50 g left and 2 minutes has elapsed.
After 2 half life you will have 25 g left and 4 minutes has elapsed.
After 3 half lives you will have 12.5 g left and 6 minutes has elapsed.
After 4 half lives you will have 6.25 g left and 8 minutes has elapsd.
After 5 half lives you will have 3.125 g left and 10 minutes has elapsed.
After 6 half lives you will have 1.5625 g left and 12 minuies has elapsed.
After 7 half lives you will have 0.78125 g left and 14 minutes has elapsed.
Or you can do it mathematically as follows:
k = 0.69315/2 min = 0.346575
ln(No/N) = kt
ln (100/N) = 0.346575*14 = 4.85205
100/N = 128
N = 100/128 = 0.78125 g left after 14 minutes and just as the experiment starts.
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