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teller

  • one year ago

Can someone tell me if this is correct? The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass. Starting with 215 grams of a radioactive isotope, how much will be left after 4 half-lives?Use the calculator provided and round your answer to the nearest gram. f=215(1+0.005)^4=219.33235763 so would it be 219.32?

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  1. teller
    • one year ago
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    @phi can you see if I got this right?

  2. phi
    • one year ago
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    The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass that means if you start with 100, after one half-life, you will have 50 your answer should be smaller than the amount you start with.

  3. teller
    • one year ago
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    Got a little confused there but would it be 210

  4. phi
    • one year ago
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    where did you get your formula? I would use 215*(0.5)^n where n is the number of half-lifes

  5. teller
    • one year ago
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    i got it from the explain tab it says to use those formulas so would it be 13.4375

  6. phi
    • one year ago
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    yes, that looks better you could do it like this 215 --> 107.5 (1 half life) 107.5/2 = 53.75 (2nd) 53.75/2 = 26.875 (3rd) 26.875/2= 13.4375 (4 half lifes) the formula is faster to use.

  7. teller
    • one year ago
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    but it said to the nearest gram so would it be 13.42?

  8. phi
    • one year ago
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    the starting amount is in grams so the answer 13.4375 is in grams round to the nearest gram means the nearest whole number you have 13.4 (plus a tiny bit more) is 13.4 closer to 13 or to 14 ?

  9. teller
    • one year ago
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    13

  10. teller
    • one year ago
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    so the answer would be 13?

  11. phi
    • one year ago
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    yes 13 the rule in rounding is look at the next lower digit 13.4 <-- look at the 4 if 5 or bigger round up (to 14) else round down to 13

  12. teller
    • one year ago
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    Thank you mind helping me with another one ?Once I get it solved since this is my last question.

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