Sodium-24, t1/2 = 15.0 hr, can be used to study the sodium balance in animals. If a saline solution containing 25 μg of 24^Na is injected into an animal, how long can the study continue before only 1 μg is left?
Show work please!
k = 0.693/t1/2
Then ln(No/N) = kt.
No = 25
N = 1
k from above.
t = solve for this.
To find out how long the study can continue before only 1 μg of Sodium-24 (24^Na) is left, we need to calculate the time it takes for half of the initial amount (25 μg) to decay. The half-life (t1/2) of Sodium-24 is given as 15.0 hours.
The decay follows an exponential decay model using the formula:
N = N0 * (1/2)^(t / t1/2)
Where:
N = amount of substance remaining after time t
N0 = initial amount of substance
t = time elapsed
t1/2 = half-life of the substance
Using the given information:
N = 1 μg (desired remaining amount, after t time)
N0 = 25 μg (initial amount)
t1/2 = 15.0 hr
Substituting these values into the formula:
1 = 25 * (1/2)^(t / 15.0)
Now, we can solve for t by solving this equation for t. Taking the natural logarithm (ln) on both sides, we have:
ln(1) = ln(25 * (1/2)^(t / 15.0))
The natural logarithm of 1 is 0, so:
0 = ln(25 * (1/2)^(t / 15.0))
Using logarithm properties, we can rewrite this equation as:
0 = ln(25) + ln((1/2)^(t / 15.0))
Applying the logarithm property ln(a^b) = b * ln(a):
0 = ln(25) + (t / 15.0) * ln(1/2)
Rearranging the equation to solve for t:
(t / 15.0) * ln(1/2) = -ln(25)
Dividing both sides by ln(1/2):
t / 15.0 = -ln(25) / ln(1/2)
Now, multiply both sides by 15.0 to isolate t:
t = 15.0 * (-ln(25) / ln(1/2))
Using a calculator to evaluate the expression on the right side, we get:
t ≈ 26.654 hours
Therefore, the study can continue for approximately 26.654 hours before only 1 μg of Sodium-24 is left.