At the start of a chemistry experiment, 50 grams of a...

Name: At the start of a chemistry experiment, 50 grams of a radioactive plutonium substance were present. After 4 hours, 3.5% of the quantity had decayed. If the rate of decay is proportional to the amount present at time t, find the amount that will remain after fifteen hours.
Uploaded: 2023-05-16T09:36:57-08:00
Duration: 5 min 35 s
Channel: Madhur L
Description: At the start of a chemistry experiment, 50 grams of a radioactive plutonium substance were present. After 4 hours, 3.5% of the quantity had decayed. If the rate of decay is proportional to the amount present at time t, find the amount that will remain after fifteen hours.

At the start of a chemistry experiment, 50 grams of a radioactive plutonium substance were present. After 4 hours, 3.5% of the quantity had decayed. If the rate of decay is proportional to the amount present at time t, find the amount that will remain after fifteen hours.

Transcript

00:01 Hi, in this question we are given with the initial amount of radioactive plutonium which is given to be let it be represented by p and this is given to be equals to 50 gram and let y represent the quantity decade after time then the remaining quantity at time t will be equals to p minus y as per the question we are given with rate of decay that is dy by dt is directly proportional to amount present at time that is the amount and solving this further we can write this as dy by dt to be equals to k times p or further as here we have the amount at time t where this is the value of the amount at time t so replacing this value we get dy by dt to be equals to k times p minus y at t and solving this further we get dy by dt plus ky is equals to kp and here finding the integrating factor to solve this we get e raised to power integration of pdx and in our case we have e raised to power integration k times dt that is equals to e raised to power kt multiplying the obtained equation with the integrating factor we get e raised to power kt times dy by dt plus k times y times e raised to power kt is equals to k times p time e raised to power kt if we check this left side this is nothing but the derivation of e raised to power kt times y this is equals to kp e raised to power kt now integrating both sides we get e raised to power kt times y is equals to k times p times integration of e raised to power kt dt and solving this further we get e raised to power kt times y is equals to kp e raised to power kt this will get cancelled now dividing both sides by e raised to power kt we get y is equals to p plus c e raised to power minus kt now using the initial condition as for t is equal to 0 we have p as equals to 50 so substituting this value we get the result as 0 to be equals to because the decayed amount initially is also 0 plus c e raised to power 0 this will give c to be equals to 50.

03:21 So we get the equation for y to be equals to p minus 50 e raised to power minus kt as we know after 4 hours 3 .5 percent of the quantity had decayed so we get y to be equals to 50 times 3 .5 by 100 and this will be equals to 1 .75 so substituting this value we get 1 .75 to be equals to 50 minus 50.

03:52 E raised to power minus k times 4 and further solving we get e raised to power minus 4 k is equals to 0 .965 or we get k to be equals to minus 1 by 4 ln 0 .965 and further solving as if we put this value of k into the equation we get y to be equals to p minus 50 e raised to power 1 by 4 ln 0 .965...

Question

At the start of a chemistry experiment, 50 grams of a radioactive plutonium substance were present. After 4 hours, 3.5% of the quantity had decayed. If the rate of decay is proportional to the amount present at time t, find the amount that will remain after fifteen hours.

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