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Math Help - Half Life

  1. #1
    Junior Member risteek's Avatar
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    Half Life

    This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

    The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

    a) What is the initial mass?
    b) What is the mass after 2 days?

    Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.
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  2. #2
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    Quote Originally Posted by risteek View Post
    This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

    The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

    a) What is the initial mass?
    b) What is the mass after 2 days?

    Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.
    y_0 = amount of S_{138} at time t = 0 days

    y = amount of S_{138} at any time, t , in days

    y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{4}}<br />

    0.64 = y_0 \left(\frac{1}{2}\right)^{\frac{12}{4}}<br />

    y_0 = 5.12 g


    you do part (b)
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by risteek View Post
    This one is half-life I know but I cannot figure out how to go through it, it isn't finding the amount so none of the examples my teacher gave me can show how to move things around...any help is appreciated.

    The half-life of sulfur-138 is 4 days. After 12 days,a sample of this radioactive material has been reduced to a mass of 0.64g.

    a) What is the initial mass?
    b) What is the mass after 2 days?

    Also if it wouldn't hurt could someone could walk through it, if not then just show all the steps to solve it.

    FormulŠ for half-life in exponential decay
    Main article: Exponential decay
    An exponential decay process can be described by any of the following three equivalent formulae:



    Nt = N0et / τ Nt = N0e − λt where
    • N0 is the initial quantity of the thing that will decay (this quantity may be measured in grams, moles, number of atoms, etc.),
    • Nt is the quantity that still remains and has not yet decayed after a time t,
    • t1 / 2 is the half-life of the decaying quantity,
    • τ is a positive number called the mean lifetime of the decaying quantity,
    • λ is a positive number called the decay constant of the decaying quantity.
    You've got N_t=0.64, and t_{1/2}=4
    and
    t=12

    so

    0.64=N_0(\frac{1}{2})^{12/4}

    What's eight times 0.64?
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  4. #4
    Junior Member risteek's Avatar
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    Thank you so much! But where does the eight come from?
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    Quote Originally Posted by risteek View Post
    Thank you so much! But where does the eight come from?
    \begin{aligned}<br />
0.64 &= N_0(\frac{1}{2})^{12/4} \\<br />
0.64 &= N_0(\frac{1}{2})^3 \\<br />
0.64 &= N_0(\frac{1}{8}) \\<br />
8(0.64) &= N_0<br />
\end{aligned}

    That's where the eight came from.


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    Quote Originally Posted by risteek View Post
    Thank you so much! But where does the eight come from?
    \left(\frac{1}{2}\right)^3 = \frac{1}{8}
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  7. #7
    Junior Member risteek's Avatar
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    Wooooow. I feel dumb I can't even use exponents right. >.< Thanks!
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    Junior Member risteek's Avatar
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    y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}<br />

    y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}<br />

    y = 2.56 g

    Is that right?
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  9. #9
    No one in Particular VonNemo19's Avatar
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    (1/2)^3=8
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    Quote Originally Posted by risteek View Post
    y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}<br />

    y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}<br />

    y = 2.56 g

    Is that right?
    Nope...
    \begin{aligned}<br />
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}} \\<br />
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} \\<br />
y &= 5.12 \left(\frac{1}{\sqrt{2}}\right) \\<br />
y &\approx 3.62<br />
\end{aligned}


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  11. #11
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    Quote Originally Posted by VonNemo19 View Post
    (1/2)^3=8
    You mean 1/8.


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  12. #12
    Junior Member risteek's Avatar
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    Quote Originally Posted by yeongil View Post
    Nope...
    \begin{aligned}<br />
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}} \\<br />
y &= 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} \\<br />
y &= 5.12 \left(\frac{1}{\sqrt{2}}\right) \\<br />
y &\approx 3.62<br />
\end{aligned}
    My teacher never taught us to square route anything with in the equation... is this question an exception?
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  13. #13
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    Quote Originally Posted by risteek View Post
    y = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}<br />

    y = 5.12 \left(\frac{1}{2}\right)^{\frac{2}{4}}<br />

    y = 2.56 g

    Is that right?
    nope ...

    y = 5.12 \left(\frac{1}{2}\right)^{\frac{1}{2}} = 3.62 g
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  14. #14
    Junior Member risteek's Avatar
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    Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys
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  15. #15
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by risteek View Post
    Oh I get it, i forgot to put the exponent into the lowest possible form. Thanks guys
    come back anytime
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