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CAN SOMEONE HELP WITH THIS QUESTION?

CAN SOMEONE HELP WITH THIS QUESTION?-example-1
User AsukaSong
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1 Answer

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Answer:

a. Since the half-life of the isotope is 8 hours, we know that the decay rate is exponential and we can use the formula:

A(t) = A0 * (1/2)^(t/8)

where A0 is the initial amount of the substance, t is the time elapsed, and A(t) is the amount of substance remaining after t hours.

Substituting the given values, we get:

A(t) = 7 * (1/2)^(t/8)

b. To find the rate at which the substance is decaying, we need to take the derivative of A(t) with respect to t:

A'(t) = -7/8 * (1/2)^(t/8) * ln(1/2)

Simplifying, we get:

A'(t) = -ln(2) * (7/8) * (1/2)^(t/8)

c. To find the rate of decay at 14 hours, we can plug in t=14 into the equation we found in part b:

A'(14) = -ln(2) * (7/8) * (1/2)^(14/8) ≈ -0.4346 grams per hour (rounded to four decimal places)

User Deafsheep
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