Isotope X has a half-life of 8 hours. How many grams will be gone if a 1.6 kg sample is left while researchers are out for 2 full days?

9 Answers

Relevance
  • Philip
    Lv 6
    1 month ago
    Favorite Answer

    Sample size at start = 1.6 kg = a, say,;

    /---------------/ after (1/3) day  =  a*(1/2)^(1);

    /------------------------/2/-----------------------/(2);

    /-------------------------3/-----------------------/(3);

    /-------------------------4/-----------------------/(4);

    /-------------------------5/-----------------------/(5);

    /-------------------------6/-----------------------/(6);

    Amount gone after 2 days = a - a*(1/2)^(6) = a[1- (1/64)] = (63/64)a;

    = (63/64)(1.6)(10^3)g = 1575g; 

  • 1 month ago

    A(t) = A(0) x (0.5)^(t/8)

    so, after 2 days we have t = 48

    Then, A(48) = 1600 x (0.5)⁶

    i.e. 25g

    :)>

  • 1 month ago

    Can't be exact, but can be estimated really close; bout 25 g. The problem with radioactive decay is while it's steady over time, it's also occurs randomly and instantaneously. Think of it like guided chaos. 

    Think of it like this.. Accuracy and precision would be like a million dog cages each with only one dog. Accuracy would be a million dog cages with a million dogs, but some cages might have 2-3 dogs, others no dogs. Precision would be the fact there are 1 million cages and 1 million dogs, this a 1:1 ratio is precise. Now say you know there are 1 million cages, 1 million dogs, but each cage can have a maximum of 2 dogs in it. How do you pick a cage at random and accurately predict how many dogs are in the cage? Now take that million and scale it to billions. While we can't precisely predict, on such a large scale, our window of accuracy scales so much it might as well be precise, even though it isn't.

  • 2 months ago

    =1.6/64=.025kg=25g

  • How do you think about the answers? You can sign in to vote the answer.
  • 2 months ago

    How many "half-lifes" are there in 2 full days. 2 days is 48 hours. 48 / 8 = 6

    So the amount of the sample is cut in half 6 times. 1.6 x (1/2)^6 = 1.6 / 64 = 0.025 kg or 25 g.

    You can do it like that even if the number of half-lifes is not a whole number, but you probably need to use a calculator to get (1/2)^x when x isn't an integer.

    Half-lifes are not hard to understand. If you reason things out like above, you don't have to worry about memorizing the formula. It will come to you automatically.

    You can do this even if what you are given isn't the half-life.  E.g., suppose you are given that 100g decays to 40 grams after 2 days. How much is left after 7 days?

    You see that you get 0.40 left after every 2 days. In 7 days there are 7/2 = 3.5 2-day periods. Just take 0.40^3.5 = 0.0405, and multiply by what you started with.

    100g x 0.0405 = 4.05 grams left. Same simple logic as the half-life case.

  • Jim
    Lv 7
    2 months ago

    A = A₀ (1/2)^(t/h)

    A = 1.6 kg (1000g/kg) (1/2)^(48/8)

    A = 1600g (1/2)^(6) g

    A = 25. g or 2.5 x 10¹g

    Attachment image
  • TomV
    Lv 7
    2 months ago

    If A₀ is the original amount, A(n) is the amount remaining after n half lives.

    A(n)/A₀ = (1/2)^n

    2 days is 48 hours

    The number of 8 hour half lives is n = 48/8 = 6

    A(6)/A₂ = (1/2)^6 = 1/64 = 0.015625

    If A₀ = 1.6 kg = 1600 g, after 2 days (6 half lives), the amount remaining will be:

    A = .0.015625*1600 = 25 g.

    I'm at a loss to explain the thumbs down received by two previous responders to this question because their answers are correct.

  • Alan
    Lv 7
    2 months ago

    2 full days = 48 hours 

    48 hours / 8 hours = 6 cycles 

    (1/2)^6  =    (1/64)  

    N = N_0*(1/2)^(t/t_half_life) =    1.6* (1/2)^(48/6)   

    so you should have 1.6 Kg *  (1/64)   = 0.025  Kg  

    or 25 grams 

  • 2 months ago

    I like to use this general equation for all growth and decay questions:

    a(t) = ae^(kt)

    Where a(t) is the amount left after "t" time

    a is the initial amount

    e is the mathematical constant

    k is the growth/decay constant

    To find a value for k, a and a(t) can be anything as long as a(t) is half of a.  So I'll use:

    a = 1

    a(t) = 0.5

    Then set t = 8 and solve for k:

    a(t) = ae^(kt)

    0.5 = e^(8k)

    ln(0.5) = 8k

    ln(0.5) / 8 = k

    Now we have a value for k so our new general equation is:

    a(t) = ae^(kt)

    a(t) = ae^(t ln(0.5) / 8)

    We can then set a = 1.6 and t = 48 (for 2 full days since our time is in hours) and solve for a(24):

    a(48) = 1.6e^(48 ln(0.5) / 8)

    a(48) = 1.6e^(6 ln(0.5))

    I'll round below to about 6 or 7 DP but won't round in my calculator to reduce errors due to rounding:

    a(48) = 1.6e^(6(-0.693147))

    a(48) = 1.6e^(-4.158883)

    a(48) = 1.6(0.015625)

    Now we're back to exact values, so no rounding is needed.

    a(48) = 0.025 kg of isotope x will remain after 2 days.

Still have questions? Get your answers by asking now.