Mathematics

Question

Iridium-192 is an isotope of iridium and has a half life of 73.83 days. If a laboratory experiment begins with 100 grams of iridium-192, the number of grams, A, of iridium-192 present after t days would be A= 100(1/2)^t/73.83. Which equation approximates the amount of iridium-192 present after t days?

2 Answer

  • Answer:

    [tex]A=100(0.990655512)^t[/tex]

    Step-by-step explanation:

    [tex]A=100(\frac{1}{2})^{\frac{t}{73.83}}[/tex]

    It does say an approximate equation so we could play with law of exponents a little especially if you don't like that fraction in the exponent.

    [tex]A=100(\frac{1}{2})^{\frac{1 \cdot t}{73.83}}[/tex]

    [tex]A=100(\frac{1}{2})^{\frac{1}{73.83}t}[/tex]

    [tex]A=100((\frac{1}{2})^{\frac{1}{73.83}})^t[/tex]

    I'm going to put (1/2)^(1/73.83) into my calculator.

    This gives me approximately: 0.990655512.

    [tex]A=100(0.990655512)^t[/tex]

    So maybe that is what you are looking for.

    Please let me know.

  • A gram of the isotope of iridium is present. Then after t day, the equation is [tex]\rm A = 100*(0.990655)^t[/tex].

    What is half-life?

    Half-Life is defined as the time required by a radioactive substance to disintegrate into a different substance. This was discovered in 1907 by Ernest Rutherford.

    Given

    Iridium-192 is an isotope of iridium and has a half-life of 73.83 days.

    [tex]\rm A= 100(0.5)^\frac{t}{73.83}[/tex]

    If A gram of the isotope of iridium is present. Then after t day, the amount will be

    [tex]\rm A= 100(0.5)^{\frac{t}{73.83}}\\[/tex]

    On simplifying, we have

    [tex]\rm A = 100((0.5)^{\frac{1}{73.83}})^{t} \\\\A = 100*(0.990655)^t[/tex]

    Thus, the required equation is [tex]\rm A = 100*(0.990655)^t[/tex].

    More about the half-life link is given below.

    https://brainly.com/question/24710827

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