00:01
In this question we are given that the radio after substance present initially mass of radioactive substance was 50 gram and after 4 hours 3 .5 % of the substance decays and we are given that the rate of decay is proportion to the amount present at time t so we have to find out the amount that will remain after 15 hours.
00:38
Here we let that y is the quantity, y is the quantity decayed after time t and we know that rate of decay is proportional to the amount of the amount of amount present at time t.
01:34
So the rate of decay that is dy over dt is proportional to the amount present at time t that is represent by pt and so if we remove the proportionality sign we have dy over dt as k times pt.
01:52
Now in place of pt we can write so pt is basically the amount present at time time t so this will be the initial amount present that is p minus the amount decayed that is y t so in place of pt we can write p minus y t so d y over d t is equal to k times p minus y where p is the initial amount and y is the amount decayed after time t so we have d y over d t as so divide over d t plus plus k y is equal to k p so here here we can write y instead of y t so it will be p minus y so k it is k p now it is a form of a differential equation in the form d y over d x plus p py equal to q.
03:07
The integrating factor is e to the bar integral p, dx and the solution of this equation is y multiplied by it is y.
03:21
The solution of the differential equation of this form is y multiplied by the integrating factor is equal to integral q multiplied by the integrating factor dx plus c.
03:34
So similarly here we find the integrating factor.
03:37
Factor that is e to the by integral in place of p we have k so k d t so we have e to the by integral e to the bar k t this is the integration factor now the solution will be y multiplied by the integrative that is e to the bar k t is equal to integral q that is k p multiplied by e to the bar k t d t plus so we have y e to the bar k t is equal to now k and p both are constant.
04:18
So it will come out of the integral and integration of e to the part k t is e to the par k t over k where c is an arbitrary constant so y multiply to e to the par k t is equal to p e to the bar k t plus c so if we make y as subject we get y is equal to p plus c e to the power minus k t.
04:48
This is our first equation.
04:52
Now we are given that at time t equal to zero the amount present initially was 50 gram and decayed was zero.
05:05
And at time t equal to four of we use this info.
05:10
So we put the values, so y is 0 is equal to 50 plus c, e to the part minus k multiplied by 0.
05:23
So it will be 0 is equal to 50 plus c, so we have 50 is equal to minus 50...