Answer:
approximately 99.53 years will pass until only 1 mg remains.
Step-by-step explanation:
To find the mass that remains after t years, we can use the formula for exponential decay:
y(t) = y(0) * (1/2)^(t / T)
where y(0) is the initial mass, T is the half-life, and y(t) is the mass remaining after t years.
In this case, y(0) = 20 mg and T = 30 years.
Plugging these values into the equation, we get:
y(t) = 20 * (1/2)^(t / 30)
This formula gives us the mass (in mg) that remains after t years.
(b) To find the mass remaining after 40 years, we plug t = 40 into the formula:
y(40) = 20 * (1/2)^(40 / 30)
Simplifying the expression:
y(40) ≈ 20 * 0.682
y(40) ≈ 13.64 mg
Therefore, approximately 13.64 mg of the sample remains after 40 years.
(c) To find the number of years it takes for only 1 mg to remain, we set y(t) equal to 1 and solve for t:
1 = 20 * (1/2)^(t / 30)
Dividing both sides by 20:
1/20 = (1/2)^(t / 30)
Taking the logarithm of both sides:
log(1/20) = log((1/2)^(t / 30))
Using the property of logarithms, we can bring down the exponent:
log(1/20) = (t / 30) * log(1/2)
Using a calculator, we can calculate the value of log(1/20) as approximately -2.9957 and log(1/2) as approximately -0.3010.
Substituting these values into the equation:
-2.9957 = (t / 30) * -0.3010
Now we can solve for t:
-2.9957 / -0.3010 = t / 30
t ≈ 99.53 years
Therefore, approximately 99.53 years will pass until only 1 mg remains.