- #1
Gregory.gags
- 31
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Question: Carbon-14 has a half-life of 5730 years. A bone is found in which [itex]\frac{7}{8}[/itex] of the original [16g of] carbon-14 has decayed. How old is the bone?
Formula: C(t)=Co([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] (it's hard to see but it says [itex]\frac{1}{2}[/itex] to the power of [itex]\frac{t}{H}[/itex])
C(t) - the number at time t
Co - the number or concentration at time 0
2-1 or ([itex]\frac{1}{2}[/itex]) - because the amount is halving
t - represents the elapsed time since measuring started (same units as H)
H - the half life
What I did was :
C(t) = [itex]\frac{1}{8}[/itex] (since [itex]\frac{7}{8}[/itex] had already decayed) so i did
[itex]\frac{1}{8}[/itex]=16([itex]\frac{1}{2}[/itex])[itex]\frac{t}{5730}[/itex]
I worked that out and got 40 110, which is wrong, but that doesn't matter right now. My problem is that the text worked it out this way:
they did everything the same as me except, after figuring the [itex]\frac{1}{8}[/itex] from the [itex]\frac{7}{8}[/itex] they multiplied it by 16 for some reason getting 2 for C(t). So they start off with 2=16([itex]\frac{1}{2}[/itex])[itex]\frac{t}{5730}[/itex] and worked it out the same way I did. I can't figure out why they did [itex]\frac{1}{8}[/itex] * 16 ? Wouldn't they have to multiply the other side of the equation by another 16 too?
They then go on to say C(t)=Co=([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] but there is no way that can be true! Can someone please explain firstly why the multiply the [itex]\frac{1}{8}[/itex] by 16 in the first place, and secondly why they equate C(t) & Co & ([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] ??
Formula: C(t)=Co([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] (it's hard to see but it says [itex]\frac{1}{2}[/itex] to the power of [itex]\frac{t}{H}[/itex])
C(t) - the number at time t
Co - the number or concentration at time 0
2-1 or ([itex]\frac{1}{2}[/itex]) - because the amount is halving
t - represents the elapsed time since measuring started (same units as H)
H - the half life
What I did was :
C(t) = [itex]\frac{1}{8}[/itex] (since [itex]\frac{7}{8}[/itex] had already decayed) so i did
[itex]\frac{1}{8}[/itex]=16([itex]\frac{1}{2}[/itex])[itex]\frac{t}{5730}[/itex]
I worked that out and got 40 110, which is wrong, but that doesn't matter right now. My problem is that the text worked it out this way:
they did everything the same as me except, after figuring the [itex]\frac{1}{8}[/itex] from the [itex]\frac{7}{8}[/itex] they multiplied it by 16 for some reason getting 2 for C(t). So they start off with 2=16([itex]\frac{1}{2}[/itex])[itex]\frac{t}{5730}[/itex] and worked it out the same way I did. I can't figure out why they did [itex]\frac{1}{8}[/itex] * 16 ? Wouldn't they have to multiply the other side of the equation by another 16 too?
They then go on to say C(t)=Co=([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] but there is no way that can be true! Can someone please explain firstly why the multiply the [itex]\frac{1}{8}[/itex] by 16 in the first place, and secondly why they equate C(t) & Co & ([itex]\frac{1}{2}[/itex])[itex]\frac{t}{H}[/itex] ??
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