Phosphorus-32 (P-32) has a half-life of 14.2 days. If 200 g of this substance are present initially, find the amount Q(t) present after t days. (Round your growth constant to four decimal places.)
How fast is the P-32 decaying when t = 21.5? (Round your answer to three decimal places.)
Well, isn't P-32 quite the attention seeker? Let's calculate its presence, shall we?
To find the amount Q(t) present after t days, we can use the equation:
Q(t) = Q(0) * 2^(-t / T)
where Q(0) is the initial amount, t is the time elapsed, and T is the half-life.
In this case, Q(0) = 200 g, T = 14.2 days, and t = 21.5 days. Plugging these values into the equation, we get:
Q(t) = 200 * 2^(-21.5 / 14.2)
Calculating, we find that Q(t) is approximately 93.9966 g.
So, after 21.5 days, there will be around 93.9966 grams of P-32 left.
Now, let's move on to the next part of the question.
To find out how fast P-32 is decaying when t = 21.5, we can use the derivative of the equation:
dQ(t) / dt = (-ln(2) / T) * Q(0) * 2^(-t / T)
Now, let's plug in the values:
dQ(t) / dt = (-ln(2) / 14.2) * 200 * 2^(-21.5 / 14.2)
Calculating this expression, we find that P-32 decays at a rate of approximately -0.287 when t = 21.5.
So, P-32 is decaying at a rate of roughly -0.287 g/day at that time.
Remember folks, decay happens, but at least P-32 is keeping things interesting!
To find the amount Q(t) of P-32 present after t days, we can use the exponential decay formula:
Q(t) = Q₀ * e^(-kt)
Where:
Q(t) is the amount of P-32 present after t days,
Q₀ is the initial amount of P-32 (200 g in this case),
k is the growth constant (we'll calculate it next).
The growth constant (k) can be found using the half-life formula:
t₁/2 = ln(2) / k
Where:
t₁/2 is the half-life of P-32 (14.2 days in this case),
ln(2) is the natural logarithm of 2.
Let's calculate k:
k = ln(2) / t₁/2
k = ln(2) / 14.2
k ≈ 0.0488 (rounded to four decimal places)
Now, we can plug the values into the exponential decay formula to find Q(t):
Q(t) = 200 * e^(-0.0488t)
To find how fast P-32 is decaying when t = 21.5 days, we need to find the derivative of Q(t) with respect to t and evaluate it at t = 21.5.
dQ(t)/dt = -0.0488 * 200 * e^(-0.0488t)
Now, substituting t = 21.5 into the derivative formula:
dQ(t)/dt = -0.0488 * 200 * e^(-0.0488 * 21.5)
dQ(t)/dt ≈ -0.0488 * 200 * e^(-1.0492)
dQ(t)/dt ≈ -0.0488 * 200 * 0.3506
dQ(t)/dt ≈ -3.426 (rounded to three decimal places)
Therefore, when t = 21.5 days, the P-32 is decaying at a rate of -3.426 g/day (rounded to three decimal places).
To find the amount Q(t) of Phosphorus-32 (P-32) present after t days, we can use the formula:
Q(t) = Q₀ * (1/2)^(t / T₀)
Where:
Q₀ = initial amount (200 g in this case)
T₀ = half-life (14.2 days in this case)
Let's calculate Q(t) after t days:
Q(t) = 200 * (1/2)^(t / 14.2)
Now, to find how fast the P-32 is decaying at t = 21.5 days, we need to find the derivative of Q(t) with respect to t and evaluate it at t = 21.5.
dQ/dt = (200 ln(1/2) / 14.2) * (1/2)^(t / 14.2)
Substituting t = 21.5:
dQ/dt = (200 ln(1/2) / 14.2) * (1/2)^(21.5 / 14.2)
We can simplify this expression to find the rate at which P-32 is decaying at t = 21.5 days.
Q(t) = 200 (1/2)^(t/14.2)
or
Q(t) = 200(.5)^(.0704t)
Q '(t) = (200/14.2) (.5)^(t/14.2)
so when t = 21.5
Q ' (t) = (200/14.2) (.5)^(21.5/14.2)
= 4.931 g/day