00:01
In this question we are given that the initial activity of a certain isotope, e .0 is given as 10 miliquries.
00:13
We are given that after some time or 4 hours, its activity has dropped to just 8 milituris.
00:24
Now by the decay law, this must be equal to the initial activity multiplied by the exponential term, e to about negative lambda t, when lambda is t, decay constant.
00:43
So given that a knot is 10, bidicuries, and we are given that the time after the initial activity is 4 hours, which we can do some manipulation, we take natural log on both sides after dividing the 10 bdicruis to the left -hand side, so just be 2 .8s.
01:09
Right and if you were to convert the four hours into seconds it would be four times three six zero right this is the total number of seconds in one hour therefore we should be able to find our lambda p .1 .55 times 10 power minus five per second from this we can also find the t half t half equals to lot 2 over lambda ndk constant substituting in ddk constant you should get should get about 4 .4 72 times 10 about 4 seconds and if you want to convert that to hours this is about 12 .4 hours now to find the total number of isotopes in the initial sample, we know that the activity is given as lambda times the total number of samples that you have.
03:19
So the initial activity is 10 mniquis.
03:24
Our decay constant is given as 1 .55 times 10 power minus 5.
03:33
We can find an n by rearranging, but we'll have to convert mini qes into backrolls, which is decay per second...