Uranium-235 has a half life of 4510 milion years. It converts to Lead-206. A sample contains 45,641 atoms of Lead-206 and 359 atoms of Uranium-235.
A) How many atoms of Uranium-235 were originally present before the sample started to decay?
B) How many years has it been since the sample started to decay?
Chemistry Half Life problem help?
2016-05-10 7:25 am
回答 (2)
2016-05-10 8:22 am
It is 238-U that decays to 206-Pb, not 235-U. 235-U decays to 207-Pb.
Furthermore, the halflife of 238-U is 4,468 million years (4.468*10^9 yrs) not 4,510 million years. The halflife of 235-U is much shorter -- 704 million years.
You cannot answer either parts (A) or (B) of this question with the information given. You need to know the amount of 206-Pb that was initially present in the sample to answer these questions.
Assuming the sample had no atoms of 206-Pb initially, and it is, in fact, a sample of initially pure 238-U (not 235-U), then the number of atoms of 238-U present initally was 45,641 + 359 = 46,000 atoms (i.e., all the 206-Pb atoms came from the decay of 238-U).
The decay equation giving the number of atoms of parent element present at ttime t (N(t)) when there were initially No atoms present is:
N(t) = No*exp(-t*ln(2)/T), where T is the halflife of the parent element.
Rearanging this to solve tor t:
lt = T*ln(No/N(t))/ln(2)
In this case (using the incorrect halflife you were given, and again assuming the sample had no 206-Pb initially):
t = (4.510*10^9 yr)*ln(46,000/45,641)/ln(2)
t = 5.098*10^7 yrs = 50.98 million years.
This is not quite accurate, because 238-U does not decay directly to 206-Pb. There is a long chain of relatively short-lived intermediate daughter products between U and Pb. The longest-lived of these is 234-U, with a halflife of 0.2455 million years. Nevertheless, the 51 million year "age" of the sample is sufficiently large relative to the 234-U halflife so that the decay series is in "secular equilibrium", and we don't really have to worry about the intermediate daughters.
This was a really poorly posed question, with several factual inaccuracies. Tell your teacher that he/she writes sloppy questions.
Furthermore, the halflife of 238-U is 4,468 million years (4.468*10^9 yrs) not 4,510 million years. The halflife of 235-U is much shorter -- 704 million years.
You cannot answer either parts (A) or (B) of this question with the information given. You need to know the amount of 206-Pb that was initially present in the sample to answer these questions.
Assuming the sample had no atoms of 206-Pb initially, and it is, in fact, a sample of initially pure 238-U (not 235-U), then the number of atoms of 238-U present initally was 45,641 + 359 = 46,000 atoms (i.e., all the 206-Pb atoms came from the decay of 238-U).
The decay equation giving the number of atoms of parent element present at ttime t (N(t)) when there were initially No atoms present is:
N(t) = No*exp(-t*ln(2)/T), where T is the halflife of the parent element.
Rearanging this to solve tor t:
lt = T*ln(No/N(t))/ln(2)
In this case (using the incorrect halflife you were given, and again assuming the sample had no 206-Pb initially):
t = (4.510*10^9 yr)*ln(46,000/45,641)/ln(2)
t = 5.098*10^7 yrs = 50.98 million years.
This is not quite accurate, because 238-U does not decay directly to 206-Pb. There is a long chain of relatively short-lived intermediate daughter products between U and Pb. The longest-lived of these is 234-U, with a halflife of 0.2455 million years. Nevertheless, the 51 million year "age" of the sample is sufficiently large relative to the 234-U halflife so that the decay series is in "secular equilibrium", and we don't really have to worry about the intermediate daughters.
This was a really poorly posed question, with several factual inaccuracies. Tell your teacher that he/she writes sloppy questions.
2016-05-10 7:40 am
A)
Originally, there are U-235 atoms but no Pb-206 atoms.
1 U-235 atom decays to give 1 Pb-206 atom.
Number of U-235 atoms originally present = 359 + 45641 = 46000
B)
Let n be the number of half-lives taken.
Fraction of U-235 atoms left :
(1/2)ⁿ = (359/46000)
log[(1/2)ⁿ] = log(359/46000)
n log(1/2) = log(359/46000)
n = log(359/46000) / log(1/2)
n = 7.00
Time taken = 4510 × 7.00 million years = 31570 million years
Originally, there are U-235 atoms but no Pb-206 atoms.
1 U-235 atom decays to give 1 Pb-206 atom.
Number of U-235 atoms originally present = 359 + 45641 = 46000
B)
Let n be the number of half-lives taken.
Fraction of U-235 atoms left :
(1/2)ⁿ = (359/46000)
log[(1/2)ⁿ] = log(359/46000)
n log(1/2) = log(359/46000)
n = log(359/46000) / log(1/2)
n = 7.00
Time taken = 4510 × 7.00 million years = 31570 million years
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