The half-life for beta decay of strontium-90 is 28.8 years. A milk sample is found to contain 10.3 ppm strontium-90. How many years would pass before the strontium-90 concentration would drop to 1.0 ppm?
k = 0.693/t1/2
Substitute k into the equation below.
ln(No/N) = kt
No = 10.3 ppm
N = 1.0 ppm
k from above.
t = ?
DR.BOB222 ENGLISH PLEASE?????? I know how to get the beta decay for strontium. just having a hard time with the concentration of strontium and the amount of yrs. passed....please help!!
All you need to do is to substitute 28.8 years into equation 1 I gave you to solve for k, then substitute k into the
ln(No/N) = kt and solve for t.
I TOLD you what to substitute for No and I told you what to substitute for N and k. You have only to solve for t. t is approximately 100 years for the Sr to decay to about 1.0 ppm.
the half-life of cobalt-60 is 5.20 years.. how many milligrams of a 2.000-mg sample remains after 6.55 years?
96.9
To determine how many years it would take for the strontium-90 concentration to drop to 1.0 ppm, we can use the concept of half-life.
First, let's understand what half-life means. The half-life of a radioactive substance is the time it takes for half of the initial amount of the substance to decay or reduce to another element. In this case, the half-life of strontium-90 is given as 28.8 years.
To find the number of half-lives it would take for the concentration to decrease to 1.0 ppm, we can use the following formula:
N = N₀ * (1/2)^(t / T)
Where:
N = Final concentration (1.0 ppm)
N₀ = Initial concentration (10.3 ppm)
t = Time passed
T = Half-life (28.8 years)
Since we're trying to solve for t, we can rearrange the formula as:
t = T * log(N / N₀) / log(1/2)
Now, let's substitute the given values into the formula:
t = 28.8 years * log(1.0 ppm / 10.3 ppm) / log(1/2)
To compute this, we can use the logarithm function on a calculator.
t ≈ 398.45 years
Therefore, it would take approximately 398.45 years for the strontium-90 concentration to drop to 1.0 ppm.