How long will it take for a 1 g sample of Polonium 210 to lose all but 1/28 of its radioactivity if the half-life is 140 d?
To determine the time it takes for a sample of Polonium 210 to lose all but 1/28 of its radioactivity, we can use the concept of half-life.
The half-life of Polonium 210 is given as 140 days. This means that every 140 days, the radioactivity of the sample will decrease by half.
To find how many half-lives are required for the sample to be reduced to 1/28 of its original radioactivity, we need to calculate the fraction remaining after each half-life until it reaches 1/28.
Let's start by finding the fraction remaining after one half-life:
Fraction Remaining = (1/2)^1 = 1/2
After the first half-life, the sample will have 1/2 of its original radioactivity remaining.
Next, we find the number of additional half-lives needed to reach 1/28:
Fraction Remaining = (1/2)^n = 1/28
To solve for 'n' (the number of half-lives), we can take the logarithm of both sides to the base 2:
log2[(1/2)^n] = log2(1/28)
n * log2(1/2) = log2(1/28)
n * (-1) = log2(1/28)
Simplifying further:
n = log2(1/28) / -1
Using a calculator, we find:
n ≈ 4.807
Since we can't have a fraction of a half-life, we round up to the nearest whole number. Therefore, it will take approximately 5 half-lives for the sample to reduce to 1/28 of its original radioactivity.
Finally, to calculate the time it will take, we multiply the number of half-lives by the half-life period:
Time = Number of half-lives * Half-life
Time = 5 * 140 days
Time ≈ 700 days
Therefore, it will take approximately 700 days for a 1 g sample of Polonium 210 to lose all but 1/28 of its radioactivity.