Nuclear Physics
U-238 decays to lead-206 by means of a series of decays.
One nucleus of U-238 eventually decays into one nucleus of lead-206
Over time the ratio of lead-206 atoms to U-238 atoms increases. This can be used to age certain types of rock.
In a particular rock sample, the ratio
number of Pb-206 atoms/number of U-238 atoms = 1/2
Show that the ratio:
number of U-238 atoms left/number of U-238 atoms initially = 2/3
Assume that the sample initially contained only U-238 atoms and subsequently it contains only U-238 atoms and lead-206 atoms.
I have no idea how to go about this so any help is greatly appreciated!
One nucleus of U-238 eventually decays into one nucleus of lead-206
Over time the ratio of lead-206 atoms to U-238 atoms increases. This can be used to age certain types of rock.
In a particular rock sample, the ratio
number of Pb-206 atoms/number of U-238 atoms = 1/2
Show that the ratio:
number of U-238 atoms left/number of U-238 atoms initially = 2/3
Assume that the sample initially contained only U-238 atoms and subsequently it contains only U-238 atoms and lead-206 atoms.
I have no idea how to go about this so any help is greatly appreciated!
Original post by Bibloski
U-238 decays to lead-206 by means of a series of decays.
One nucleus of U-238 eventually decays into one nucleus of lead-206
Over time the ratio of lead-206 atoms to U-238 atoms increases. This can be used to age certain types of rock.
In a particular rock sample, the ratio
number of Pb-206 atoms/number of U-238 atoms = 1/2
Show that the ratio:
number of U-238 atoms left/number of U-238 atoms initially = 2/3
Assume that the sample initially contained only U-238 atoms and subsequently it contains only U-238 atoms and lead-206 atoms.
I have no idea how to go about this so any help is greatly appreciated!
One nucleus of U-238 eventually decays into one nucleus of lead-206
Over time the ratio of lead-206 atoms to U-238 atoms increases. This can be used to age certain types of rock.
In a particular rock sample, the ratio
number of Pb-206 atoms/number of U-238 atoms = 1/2
Show that the ratio:
number of U-238 atoms left/number of U-238 atoms initially = 2/3
Assume that the sample initially contained only U-238 atoms and subsequently it contains only U-238 atoms and lead-206 atoms.
I have no idea how to go about this so any help is greatly appreciated!
You must read and digest all of the information presented before you can answer this. It is a case of logic and ratios:
The question tells you the sample initially (originally) contained only U238 atoms.
One U238 decays to one Pb-206.
The rock sample now contains Pb-206 / U-238 in the ratio of 1/2 i.e. 3 parts, all of which must have originally been uranium but now 1 part is lead and 2 parts are still uranium.
If there are still 2 parts of U-238 left and there were 3 initial parts, then the ratio between U-238 left / U-238 initial must be 2 parts / 3 parts (2/3).
(edited 9 years ago)
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