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Half-life question watch

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    The radioactive isotopes, magnesium-25 has a half-life of 17 hrs and and argon-40 has a half-life of 28.2 hrs. A mixture has the same amount of magnesium-25 and argon-40 atoms at the the start. Find out the time which the number of magnesium-25 atoms becomes half the number of argon-40 atoms.

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    Demonstrate that when the time increases, the ratio of the two becomes smaller and approaches zero as t tends to infinity.

    No idea how to do this part.
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    (Original post by Airess3)
    The radioactive isotopes, magnesium-25 has a half-life of 17 hrs and and argon-40 has a half-life of 28.2 hrs. A mixture has the same amount of magnesium-25 and argon-40 atoms at the the start. Find out the time which the number of magnesium-25 atoms becomes half the number of argon-40 atoms.

    Name:  20151111_232434.jpg
Views: 51
Size:  403.2 KB

    Demonstrate that when the time increases, the ratio of the two becomes smaller and approaches zero as t tends to infinity.

    No idea how to do this part.
    I think it would be easier to work out the time if you take the ratio of the number of both elements. Although your reasoning is fine, it wouldn't quite be correct to write: 1/2Mg e-at =Ar e-at . This is because the equation indicates that at the time t the number Ar is half the number of Mg atoms, and not twice, because 1/2Mg e-atis equal to the number of Ar atoms.
    As with considering their rations, you know that at a certain time the number of Mg is half of the number of Ar atoms, or the number of Ar is twice the number of Mg atoms. Let's for simplicity represent the decay constant of each element with their corresponding symbol, i.e aMg=Mg.
    Therefore (Ar e-Art)/(Mg e-Mgt)=2
    And so 2Mg e-Mgt=Ar e-Art ----------> 2 e-Mgt= e-Art

    Taking natural logs of both sides results in: ln(2)-Mgt=-Art , or Art-Mgt=ln(2)
    Solvinf for t: t=[ln(2)]/[Ar-Mg]
    Since you have the half life of both elements, you can work out their decay constant and find t. I'm not sure about the second part where you need to derive an equation as to show the variation of the ratio of the number of both elements with respect to time.
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    (Original post by Mehrdad jafari)
    I think it would be easier to work out the time if you take the ratio of the number of both elements. Although your reasoning is fine, it wouldn't quite be correct to write: 1/2Mg e-at =Ar e-at . This is because the equation indicates that at the time t the number Ar is half the number of Mg atoms, and not twice, because 1/2Mg e-atis equal to the number of Ar atoms.
    As with considering their rations, you know that at a certain time the number of Mg is half of the number of Ar atoms, or the number of Ar is twice the number of Mg atoms. Let's for simplicity represent the decay constant of each element with their corresponding symbol, i.e aMg=Mg.
    Therefore (Ar e-Art)/(Mg e-Mgt)=2
    And so 2Mg e-Mgt=Ar e-Art ----------> 2 e-Mgt= e-Art

    Taking natural logs of both sides results in: ln(2)-Mgt=-Art , or Art-Mgt=ln(2)
    Solvinf for t: t=[ln(2)]/[Ar-Mg]
    Since you have the half life of both elements, you can work out their decay constant and find t. I'm not sure about the second part where you need to derive an equation as to show the variation of the ratio of the number of both elements with respect to time.
    Your reasoning is correct, but your use of symbols is confusing(use of the same symbol for different quantities). We know the initial number of nuclei is the same for each nuclide, so you can write this as N_0 in each case.
    (Original post by Airess3)
    The radioactive isotopes, magnesium-25 has a half-life of 17 hrs and and argon-40 has a half-life of 28.2 hrs. A mixture has the same amount of magnesium-25 and argon-40 atoms at the the start. Find out the time which the number of magnesium-25 atoms becomes half the number of argon-40 atoms.

    Name:  20151111_232434.jpg
Views: 51
Size:  403.2 KB

    Demonstrate that when the time increases, the ratio of the two becomes smaller and approaches zero as t tends to infinity.

    No idea how to do this part.
    For the second part, write the ratio in terms of what you know and t(like you did for the 1st part, but with the ratio unknown), then take limits.
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    (Original post by morgan8002)
    Your reasoning is correct, but your use of symbols is confusing(use of the same symbol for different quantities). We know the initial number of nuclei is the same for each nuclide, so you can write this as N_0 in each case.


    For the second part, write the ratio in terms of what you know and t(like you did for the 1st part, but with the ratio unknown), then take limits.
    True, I believe at least I should have represented the number of nuclei of each element with N_0 or Nthe element
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    (Original post by morgan8002)
    Your reasoning is correct, but your use of symbols is confusing(use of the same symbol for different quantities). We know the initial number of nuclei is the same for each nuclide, so you can write this as N_0 in each case.


    For the second part, write the ratio in terms of what you know and t(like you did for the 1st part, but with the ratio unknown), then take limits.
    Just to clarify, so do I take the limits of [-aMgT ] + [aArT] with the numbers substituted in. Or do I also need to include the 'e' in both and make it 'to the power'?
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    (Original post by Airess3)
    Just to clarify, so do I take the limits of [-aMgT ] + [aArT] with the numbers substituted in. Or do I also need to include the 'e' in both and make it 'to the power'?
    You haven't proved that the exponential of a limit is equal to the limit of the exponential. I think this is true but it's best not to use results you haven't proven.
    Take the limit of the ratio of the number of atoms of each type.
    Also, don't take the limits separately. The quotient rule for limits only works if the limit of the denominator is non-zero. In this case, both limits are zero.
 
 
 
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