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The radioisotope strontium-90 has a half-life of 38.1 years. If a sample contains 100 mg of Sr-90, how many milligrams will remain when the age of the sample is 190.5 years?

Solution:

t_{1/2} = 38.1 years, t = 190.5 years
From the equation t_{1/2} = 0.693/λ, we obtain
λ = 0.693/t_{1/2} = 0.693/38.1 yr = 0.018189 yr^{-1}

Answer :

To determine how much of the original 100 mg of strontium-90 ([tex]^{90}Sr[/tex]) will remain after 190.5 years, we can use the concept of radioactive decay, specifically the half-life formula.

Key Information:


  • Half-life ([tex]t_{1/2}[/tex]): 38.1 years

  • Initial amount: 100 mg

  • Elapsed time ([tex]t[/tex]): 190.5 years


Step-by-step Solution:


  1. Determine the Decay Constant ([tex]\lambda[/tex]):

    The decay constant ([tex]\lambda[/tex]) can be calculated using the half-life equation:

    [tex]\lambda = \frac{0.693}{t_{1/2}}[/tex]

    Plug in the values:

    [tex]\lambda = \frac{0.693}{38.1} \approx 0.018189 \text{ year}^{-1}[/tex]


  2. Use the Exponential Decay Formula:

    The formula for exponential decay is:

    [tex]N(t) = N_0 \times e^{-\lambda t}[/tex]

    Where:


    • [tex]N(t)[/tex] is the amount remaining after time [tex]t[/tex].

    • [tex]N_0[/tex] is the initial amount.


    Substituting the known values:

    [tex]N(190.5) = 100 \times e^{-0.018189 \times 190.5}[/tex]

    Calculate the exponent:

    [tex]-0.018189 \times 190.5 \approx -3.4613[/tex]

    Calculate the remaining amount using the exponential function:

    [tex]N(190.5) = 100 \times e^{-3.4613} \approx 100 \times 0.031[/tex]

    Therefore:

    [tex]N(190.5) \approx 3.1 \text{ mg}[/tex]



Conclusion:

After 190.5 years, approximately 3.1 milligrams of the original 100 mg sample of strontium-90 will remain.

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