00:01
For this problem on the topic of nuclear physics, we are told that uranium 238, which is the most stable isotope of uranium, has an alpha decay half -life of 4 .5 times 10 to the 9 years.
00:14
Plotonium's more stable isotope, plutonium 244 has a half -life of 8 times 10 to the 7 years, and curium 248 has a half -life of 3 .4 times 10 to the 5 years.
00:26
When half an original sample of 238 has decayed, we want to know the fraction of the 1 .4.
00:31
Of the original sample of plutonium and curiam that is left.
00:37
Now the fraction of undecayed nuclear remaining after a time t is given by n over n0, and this is equal to e to the minus lambda t, which is e to the minus natural log of 2 divided by the half -life of the element t -half.
00:58
Lambda here is the disintegration constant and t half is the natural log of 2 over lambda so for plutonium 244 at t is equal to 4 .5 times 10 to the 9 years we have lambda t is equal to the natural log of 2 times the time t divided by the half -life t half and this is the natural log of 2 times 4 .5 times 10 to the 9 years, which is the half -life of uranium over the half -life for plutonium 8 times 10 to the 7 years, which gives us 39.
01:53
And so therefore the fraction remaining n over n -0 is e to the minus 39, which is, 1 .2 times 10 to the minus 17...