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

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. Reply

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 N12

=

N

0

/4096 or 1/4096.