If Radium-223 has a half life of 10.33 days, what time duration would it require for the activity associated with this sample to decrease 1.5% of its present value?
To find the time duration required for the activity of Radium-223 to decrease by 1.5% of its present value, we can use the formula for radioactive decay:
A(t) = A₀ * e^(-kt)
Where:
A(t) is the activity at time t
A₀ is the initial activity (present value)
k is the decay constant
t is the time duration
We need to find the time duration when the activity decreases by 1.5% of its present value. Mathematically, this can be represented as:
A(t) = A₀ - (0.015 * A₀)
Simplifying the equation:
A₀ * e^(-kt) = A₀ - (0.015 * A₀)
Dividing both sides by A₀:
e^(-kt) = 1 - 0.015
Taking the natural logarithm (ln) of both sides:
-kt = ln(1 - 0.015)
Simplifying further:
t = -ln(1 - 0.015) / k
Given that the half-life (T₁/₂) of Radium-223 is 10.33 days, we can determine the decay constant (k) by using the formula:
k = (ln(2)) / T₁/₂
Substituting the value of T₁/₂:
k = (ln(2)) / 10.33
Now we can substitute this value of k into the equation for t:
t = -ln(1 - 0.015) / [(ln(2)) / 10.33]
Calculating this expression would give us the time duration required for the activity of Radium-223 to decrease by 1.5% of its present value.
To determine the time duration required for the activity associated with Radium-223 to decrease by 1.5% of its present value, we can use the concept of half-life.
The formula to calculate the remaining fraction of a substance after a certain number of half-lives is:
Remaining fraction = (1/2)^(number of half-lives)
Given that the half-life of Radium-223 is 10.33 days, we can calculate the number of half-lives required for the activity to decrease by 1.5%:
1.5% = 0.015 (representing 1.5/100)
Remaining fraction = 1 - 0.015 = 0.985 (representing 98.5% of the present value)
Now, we can set up the equation and solve for the number of half-lives:
0.985 = (1/2)^(number of half-lives)
Taking the logarithm (base 2) of both sides of the equation:
log2(0.985) = number of half-lives
Using a scientific calculator, we can find that log2(0.985) is approximately -0.0253.
Therefore, the number of half-lives is approximately -0.0253.
To convert the number of half-lives into a time duration, we need to multiply it by the half-life of Radium-223:
time duration = number of half-lives * half-life
time duration = -0.0253 * 10.33
Solving this equation, we find that the time duration required for the activity associated with Radium-223 to decrease by 1.5% of its present value is approximately -0.26 days.
It is important to note that a negative time duration doesn't make sense in this context since time cannot be negative. It is possible that there may be an error in the initial question or calculation.
(1/2)^(x/10.33) = 0.985
x = 0.225 days
Now, if you meant to say "decrease to 1.5%" then that would be
(1/2)^(x/10.33) = 0.015
x = 68.63 days