Question

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

Answer 1

The correct answer is - 1/4096.

Step-by-step 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.