00:01
Hello everyone we need to find how many grams of plutonium 239 is present after 25 ,000 years.
00:07
So let us consider the function as a is equal to a0 into e power k t.
00:16
So since this is the problem of decay of plutonium 239, we have considered a function as an exponential decay function.
00:26
So clearly from this we can write a is equal to 16 e power k t according to the given question and we are given with t is equal to 25 000 so first let us compute a which is nothing but after 25 000 years a will be equal to 16 by 2 which is equal to 8 so using this initial value of a we can find the value of k that is 8 is equal to 16 e power 25 ,000 k or we can write this as 0 .5 is equal to e power 25 ,000 k.
01:21
Now taking log on both sides, we would get log 0 .5 is equal to 25 ,000k.
01:30
Or the value of this is minus 0 .000000.
01:37
Double 7 3 that is the value of k so re -substuting the value of k in this equation we would get a is equal to 16 e power minus 0 .0000273 so let us name this equation as 1 now we need to find the amount of plutonium 239 present after 50 ,000 ,000 ,000, so for that we need to substitute t is equal to 50 ,000 in this equation 1.
02:23
So we would obtain a is equal to 16 e power minus 0 .0000 -2773 into 50 ,000.
02:41
So the value of a will be approximately equal to 4 ,000.
02:47
Grams similarly to find the amount of plutonium 239 present after 75 ,000 years we take t is equal to 75 ,000 and substitute in equation 1 we would get a is equal to 16 e power minus 0 .0000273 into 75 ,000 so further simplifying this we we would get the value of a to be approximately equal to 2 grams...