Radioactive 24Na with a half-life of 14.8 hours is injected into an animal in a tracer experiment. How many hours will it take for the radioactivity to fall to 12.5% of its original intensity?

Question

Radioactive 24Na with a half-life of 14.8 hours is injected into an animal in a tracer experiment. How many hours will it take for the radioactivity to fall to 12.5% of its original intensity?

I am confused on what number to start with

Answer 1

You have the half life, Use that to calculate k.

k = 0.693/t_1/2

Then
ln(No/N) = kt
Use 100 for No
Use 12.5 for N
Substitute k from above.
Solve for t in hours if you use t_1/2 in hours.

Answer 2

To solve this problem, we need to start with the initial activity of the radioactive substance.

Let's assume the initial activity is represented by A₀.

At time t = 0, the initial intensity is 100% or 1.0. So, we can write:
A₀ = 1.0

Now, we need to calculate the time it takes for the radioactivity to fall to 12.5% of its original intensity, which means it will be 0.125 times the initial intensity.

Using the formula for radioactive decay, we can relate the time (t) and the remaining intensity (I):
I = A₀ * (1/2)^(t/h)

Where:
- I is the remaining intensity
- A₀ is the initial intensity
- t is the time elapsed
- h is the half-life of the substance

We know that at this given time, the remaining intensity is 0.125 times the initial intensity. So we can rewrite the equation as:
0.125 = 1.0 * (1/2)^(t/14.8)

To solve for t, we can take the logarithm (base 2) of both sides:
log₂(0.125) = log₂((1/2)^(t/14.8))

Using the properties of logarithms, we can bring down the exponent:
log₂(0.125) = (t/14.8) * log₂(1/2)

Now, we can solve for t:
t/14.8 = log₂(0.125) / log₂(1/2)
t = (log₂(0.125) / log₂(1/2)) * 14.8

Using a calculator, we find:
t ≈ 49.07 hours

Therefore, it will take approximately 49.07 hours for the radioactivity to fall to 12.5% of its original intensity.

Answer 3

To find out how many hours it will take for the radioactivity to fall to 12.5% of its original intensity, we first need to determine the fraction of the original intensity remaining after a certain time based on the half-life of the radioactive substance.

The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, the half-life of 24Na is 14.8 hours. This means that after 14.8 hours, the intensity of the radioactive substance will be reduced by half.

To solve the problem, we can use the formula:

Remaining intensity = Initial intensity * (1/2)^(time / half-life)

Let's plug in the values we know:
Remaining intensity = 12.5% = 0.125 (since 12.5% is equal to 0.125 as a decimal)
Initial intensity = 100% = 1 (since we are measuring the remaining intensity as a fraction of the initial intensity)
Half-life = 14.8 hours

Now, we can rearrange the formula and solve for time:
0.125 = 1 * (1/2)^(time / 14.8)

To solve for time, we can take the logarithm of both sides of the equation (to the base 1/2) to isolate the exponent:

log(base 1/2) 0.125 = log(base 1/2) [(1/2)^(time / 14.8)]

Using the logarithmic identity log(base a) b^c = c * log(base a) b, the equation becomes:

time / 14.8 = log(base 1/2) 0.125

Now, we can solve for time by multiplying both sides by 14.8:

time = 14.8 * log(base 1/2) 0.125

Using a calculator, the value of log(base 1/2) 0.125 is approximately -2.66.

Therefore, time ≈ 14.8 * -2.66 ≈ -39.43 hours.

Since time cannot be negative, we can ignore the negative sign. Thus, it will take approximately 39.43 hours (or rounded to 39 hours) for the radioactivity to fall to 12.5% of its original intensity.