00:01
So the half life of iodine is given us 8 .065 days and that is equivalent to 193 .5 6 hours.
00:08
So here the general equation is given by n of t equals n0 e to the power minus k t.
00:16
Here we have n t to be half of the n knot.
00:19
So when we solve this, what are we going to get? we have the natural log of 2 kt and this is because is for the half -life so we have equals kt and the half -life of the harylady are you doing sorry so from here we have t a half -life because length all over and that is equal to 1 930 points by six to here the key becomes equal to three points three points 5 8 times 10 to the power minus 3 to the power minus 3 okay now we should know that the bottle of iodine bottle of iodine 1 3 1 bottle of iodine 131 2 is delivered to the hospital on 5th september it delivered to hospital on 5th september and the radio activity at 12 p .m on september 10th is given as 380 mega macros mega breadcross so the general equation that we have nt equals n0 e to the par minus k t we assume t equals zero on september and um 12 p m september 10 we have 380 for the nt our time becomes 120 hours then the k become 3 .5 8 and since the par minus 3 if that is the case then this what we have you have 380 equals n knots e to the power minus k t and that is 120 multiplied by 3 .5 8 times 10 to the power minus so we have n knots equals 380 all over 0 .6507 and that is equivalent or that is equal to 612 .612 .612 2 .2 .2 .21 mega bercule.
03:38
So just to highlight this well, 612 612 .2 .2 .2 .2 .2 .2 .2 .2 .2 1 .2 .2 1 2 1 2 1 2 1 2 1 mega mega okay so the amount of iodine one three one on september 5th at 12 p .m.
04:08
Is given to us a 6 .1 2 .2 .2 p.
04:13
2 1 mega macro so 6 a .m.
04:21
On september 25th 15 .4 .0 that's at 12 p .m.
04:27
Also on september 5th then we have this the time becomes 4 .7 4 hours or seven four hours now we can go back to the general equation and look at our n of t we'll go back to our general equation we have n of t to be equal to 6 1 2 .21 multiplied by the exponent e to the power minus 474 k t so multiplied by our k 0 .0 .0.
05:14
0375 and this gives us the n t as 1 .03 points 103 .5 mega macro.
05:46
Okay and this is the n of the t.
05:51
Now let's look at the third one.
05:53
Let's look at the third one.
05:53
Let's look at the third one.
05:54
With that, we have the general equation here...