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anonymous

  • 5 years ago

The half-life of a certain radioactive element is 10 years. How much of a 20g sample will be left after 8 years? (Hint: first find the decay constant.)

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  1. anonymous
    • 5 years ago
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    \[ N(t) = N_0 e^{-\lambda t}. \, \] use this, substitute the values in, and you will get the decay constant

  2. anonymous
    • 5 years ago
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    I already used 10 = ln(2) / k ---> ln(2) / 10 = k. But then what?

  3. anonymous
    • 5 years ago
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    Then I got A = 20e^((ln(2)/10)10). What's next? where do the 8 years come in?

  4. anonymous
    • 5 years ago
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    ok, the amount of radioactive element left after 8 years will be amount remained =20/(2(8/10)) =12.5g

  5. anonymous
    • 5 years ago
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    in my oppinion

  6. anonymous
    • 5 years ago
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    where'd you get the 20 from?

  7. anonymous
    • 5 years ago
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    oh yea i see sorry.

  8. anonymous
    • 5 years ago
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    ok, what does half-life mean in the first pace? it is 10 years, meaning that after 10 years the amount will become smaler by a factor of two, that is we need to divide it by 2, but since we only wait 8 years, which is 8/10 or 0.8 of the half life,we have to multiply the half life by the ratio 0.8 to 1 so, it is 2*0.8=1.6, so after 8 years, the sample reduses by a factor of 1.6

  9. dumbcow
    • 5 years ago
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    Use the equation posted at the top and use t=8, k = ln(2)/10, N_0 = 20 Answer should be around 11.49 rounded

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