Suppose f(t) = P0e^-0.3t grams of radioactive substance are present after t seconds. If 200 grams of substance are present initially, how much is present after 4 seconds? What is the half-life of the substance?
f(4) = 200 e^(-.3(4))
= ...
for 2nd part:
half of 200 is 100
100 = 200 e^(-.3t)
.5 = e^(-.3t)
take ln of both sides
ln .5 = -.3t lne , remember lne = 1
etc
To find the amount of substance present after a given time, we can substitute the value of t into the equation for f(t) and solve it.
Given:
f(t) = P0 * e^(-0.3t)
P0 = 200 grams (initial amount)
t = 4 seconds
To find the amount of substance present after 4 seconds, substitute the values into the equation:
f(4) = 200 * e^(-0.3 * 4)
f(4) = 200 * e^(-1.2)
f(4) ≈ 200 * 0.30119421191
f(4) ≈ 60.238842381
Therefore, approximately 60.24 grams of the substance are present after 4 seconds.
To find the half-life of the substance, we need to find the value of t when f(t) is equal to half of the initial amount (P0/2).
P0/2 = P0 * e^(-0.3t)
0.5 = e^(-0.3t)
To solve for t, take the natural logarithm (ln) of both sides:
ln(0.5) = ln(e^(-0.3t))
ln(0.5) = -0.3t * ln(e)
ln(0.5) = -0.3t
Now divide both sides by -0.3:
t = ln(0.5) / -0.3
Using a calculator, you can find the value of t to be approximately 2.31 seconds.
Therefore, the half-life of the substance is approximately 2.31 seconds.