In eight days, a certain amount of Vanadium-48, V^48, decays to 1 over square root 2 of its original amount. What is the half life of Vanadium-48?
P*V = k
200*32 = V*40
that would be 16 days, right?
1/√2 * 1/√2 = 1/2
To find the half-life of Vanadium-48, we need to understand the relationship between the amount of the substance and time.
Let's assume that the original amount of Vanadium-48 is represented by A0. After eight days, the amount of Vanadium-48 remaining is 1/√2 times the original amount. We can represent this as:
A8 = (1/√2) * A0
Now we need to find the time it takes for half of the substance to decay. The half-life is the amount of time for the remaining amount to become half of the original amount. In this case, we need to find the value of t where:
A0/2 = A8
Let's substitute the values we have:
A0/2 = (1/√2) * A0
To solve for A0/2, we can multiply both sides of the equation by √2 to eliminate the fraction:
√2 * (A0/2) = √2 * (1/√2) * A0
√2 * (A0/2) = A0
Now, let's simplify the equation:
√2 * (A0/2) = A0
√2 * A0/2 = A0/2
A0 * √2/2 = A0/2
√2/2 = 1/2
Now that we have found that √2/2 = 1/2, we can conclude that the half-life of Vanadium-48 is 8 days, because after 8 days, half of the substance will remain.
Therefore, the half-life of Vanadium-48 is 8 days.