jeffrey629
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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?



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