+12 Consider the radioactive decay formula A=Aoe^-kt where a is the amount of radium remaining at the time t. Ao is the amount present initially and k is decay constant. After How many years would 10 grams of radium decay so that only 8 grams remain? The half life of radium is 1590 years
+130073 We have5 grams remaining after 1590 yeats.....so.....
5 = 10 e^[-k(1590)] divide both sides by 10
(.5) = e^[-k(1590] take the ln of both sides
ln (.5) = ln e^[-k(1590] and we can write
ln (.5) = [-k(1590] *ln e and ln e = 1....so we have
ln(.5) = -k(1590) divide both sides by 1590
ln(.5)/1590 = -k
-0.0004359416229937 = -k .....so.....k = 0.0004359416229937
And we need to find t for 8 grams remaining....so.....
8 = 10 e^[-0.0004359416229937 (t)] divide both sides by 10
.8 = e^[-0.0004359416229937 (t)] take the ln of both sides
ln(.8) = ln e^[-0.0004359416229937 (t)] and we can write
ln(.8) = [-0.0004359416229937 (t)] ln e and ln e = 1 .....so we have...
ln(.8) = -0.0004359416229937 (t) divide both sides by -0.0004359416229937
ln(.8) / -0.0004359416229937 = t = about 511.87 years
+130073 We have5 grams remaining after 1590 yeats.....so.....
5 = 10 e^[-k(1590)] divide both sides by 10
(.5) = e^[-k(1590] take the ln of both sides
ln (.5) = ln e^[-k(1590] and we can write
ln (.5) = [-k(1590] *ln e and ln e = 1....so we have
ln(.5) = -k(1590) divide both sides by 1590
ln(.5)/1590 = -k
-0.0004359416229937 = -k .....so.....k = 0.0004359416229937
And we need to find t for 8 grams remaining....so.....
8 = 10 e^[-0.0004359416229937 (t)] divide both sides by 10
.8 = e^[-0.0004359416229937 (t)] take the ln of both sides
ln(.8) = ln e^[-0.0004359416229937 (t)] and we can write
ln(.8) = [-0.0004359416229937 (t)] ln e and ln e = 1 .....so we have...
ln(.8) = -0.0004359416229937 (t) divide both sides by -0.0004359416229937
ln(.8) / -0.0004359416229937 = t = about 511.87 years
