Question

Phosphorus-33 has a half-life of 25 days. What fraction of the original sample would remain after 300 days?

Answer 1

Answer:

The correct answer is - 1/4096.

Explanation:

In the starting, there is that the number of atoms is  N 0 .

The number of half-lives in this case, = 300/ 25 = 12

After the first half-lives or 25 days,  N 1 = N 0 /2

the half life is that time where half of the sample had decay.

After the second half-life or 50 days,  N 2 = N 1 /2  = N 0 /4 .

There is division by two the original amount, then after four times you divide four times for  2  that means that you divide by  2 ^12  =4096 .

So the final amount remain is  N 12 = N 0 /4096  or 1/4096.