The half-life of radioactive strontium-90 is approximately 31 years. In 1960, radioactive strontium-90 was released into the atmosphere during testing of nuclear weapons, and was absorbed into people's bones. How many years does it take until only 11 percent of the original is left?
.11=e^(-ln2*years/31)
take the ln of each side..
ln(.11)=ln2 * years/31
years= -31*ln(0.11)/ln(2)= ????
To determine how many years it takes until only 11 percent of the original radioactive strontium-90 is left, we can use the concept of half-life.
The half-life of strontium-90 is approximately 31 years, meaning that every 31 years, the amount of strontium-90 reduces by half.
To find out how many half-lives it takes until only 11 percent remains, we can use the following formula:
Number of half-lives = ln(Percent remaining) / ln(0.5)
Where ln represents the natural logarithm and ln(0.5) is equal to -0.693.
Let's calculate the number of half-lives:
Number of half-lives = ln(0.11) / ln(0.5)
Number of half-lives ≈ -2.197 / -0.693
Number of half-lives ≈ 3.17
Since we cannot have a fraction of a half-life, we will round up to the nearest whole number. Therefore, it takes approximately 4 half-lives for only 11 percent of the original strontium-90 to remain.
To find how many years this is, we can multiply the number of half-lives by the length of a half-life (31 years):
Total years = Number of half-lives x Half-life
Total years = 4 x 31
Total years ≈ 124 years
Therefore, it takes approximately 124 years until only 11 percent of the original radioactive strontium-90 is left.
To determine the number of years it takes until only 11 percent of the original radioactive strontium-90 is left, we can use the concept of the half-life.
The half-life of a radioactive substance is the time it takes for half of its original quantity to decay or transform into a different element. In this case, the half-life of strontium-90 is approximately 31 years.
To calculate the years until only 11 percent of the original is left, we can use the following equation:
Final amount = Initial amount × (1/2)^(n)
Where "Final amount" is the remaining amount of strontium-90, "Initial amount" is the original amount of strontium-90, and "n" is the number of half-lives.
In this case, we want to find the number of years it takes until only 11 percent (0.11) of the original amount is left. So, we can substitute these values into the equation:
0.11 = 1 × (1/2)^(n)
Now we need to solve for "n," the number of half-lives. Rearranging the equation, we have:
(1/2)^(n) = 0.11
To isolate the exponential term, we can take the logarithm (base 1/2) of both sides:
log base (1/2) (1/2)^(n) = log base (1/2) (0.11)
n = log base (1/2) (0.11)
Using logarithmic properties, we can convert the base 1/2 to base 10:
n = log base (10) (0.11) / log base (10) (1/2)
Calculating this using a calculator, we find:
n ≈ 5.24
Therefore, it would take approximately 5.24 half-lives for only 11 percent of the original radioactive strontium-90 to remain.
Now, to find the number of years, we multiply the number of half-lives by the half-life of strontium-90:
Number of years = 5.24 × 31 years
Calculating this multiplication, we find:
Number of years ≈ 162.44 years
Thus, it would take approximately 162.44 years until only 11 percent of the original radioactive strontium-90 is left.