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 N
12
=
N
0
/4096 or 1/4096.