To solve this problem, we need to use the formula for radioactive decay which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity,
N0 is the initial quantity,
t is the time,
T is the half-life.

We are given:
N0 = 8.80mg,
N = 5.40mg,
T = 8.04 days.

We need to find t.

Substituting the given values into the formula, we get:

5.40 = 8.80 * (1/2)^(t/8.04)

To isolate t, we first divide both sides by 8.80:

5.40/8.80 = (1/2)^(t/8.04)

Then, take the natural logarithm (ln) of both sides:

ln(5.40/8.80) = ln((1/2)^(t/8.04))

Using the property of logarithms that allows us to bring down exponents, we get:

ln(5.40/8.80) = (t/8.04) * ln(1/2)

Finally, solve for t:

t = 8.04 * ln(5.40/8.80) / ln(1/2)

Calculating the right side, we get:

t ≈ 5.60 days

So, it would take approximately 5.60 days for the sample to decay from 8.80mg to 5.40mg.

Question

To solve this problem, we need to use the formula for radioactive decay which is:

N = N0 * (1/2)^(t/T)

where:
N is the final quantity,
N0 is the initial quantity,
t is the time,
T is the half-life.

We are given:
N0 = 8.80mg,
N = 5.40mg,
T = 8.04 days.

We need to find t.

Substituting the given values into the formula, we get:

5.40 = 8.80 * (1/2)^(t/8.04)

To isolate t, we first divide both sides by 8.80:

5.40/8.80 = (1/2)^(t/8.04)

Then, take the natural logarithm (ln) of both sides:

ln(5.40/8.80) = ln((1/2)^(t/8.04))

Using the property of logarithms that allows us to bring down exponents, we get:

ln(5.40/8.80) = (t/8.04) * ln(1/2)

Finally, solve for t:

t = 8.04 * ln(5.40/8.80) / ln(1/2)

Calculating the right side, we get:

t ≈ 5.60 days

So, it would take approximately 5.60 days for the sample to decay from 8.80mg to 5.40mg.

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