The half-life of a radio-active element is 9 days, what fraction of atoms has decayed in 36 days?
\( \frac{1}{16} \)
\( \frac{1}{4} \)
\( \frac{1}{2} \)
\( \frac{15}{16} \)
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Explanation:
T(1/2) = 9 days
t = 36days
NR = No, (1/2)n and T(1/2) = t/n
For fraction No = 1
but for percentage No = 100
T(1/2) = t/n
9 = 36/n
n = 36/9 = 4
NR = 1(1/2)4
NR = 1/16
The fraction of atom remaining is 1/16
The fraction of atom that have decay = No - NR = 1 - 1/16 = 15/16
it takes 9 days for the substance to deacay to half. So in 36 days it means it decayed four times.
x - x - x - x - x
so it is 1/16 of the mass there
Let T
1/2
be the half-life of the radioactive element, which is given as 9 days.
Let t be the total time elapsed, which is given as 36 days.
The number of half-lives that have occurred in time t is given by:
n=
T
1/2
t
=
9 days
36 days
=4
After each half-life, half of the remaining radioactive atoms decay. Let N
0
be the initial number of atoms.
After 1 half-life (9 days), the number of remaining atoms is N
1
=N
0
×(
2
1
)
1
=
2
1
N
0
. The fraction decayed is 1−
2
1
=
2
1
.
After 2 half-lives (18 days), the number of remaining atoms is N
2
=N
0
×(
2
1
)
2
=
4
1
N
0
. The fraction decayed is 1−
4
1
=
4
3
.
After 3 half-lives (27 days), the number of remaining atoms is N
3
=N
0
×(
2
1
)
3
=
8
1
N
0
. The fraction decayed is 1−
8
1
=
8
7
.
After 4 half-lives (36 days), the number of remaining atoms is N
4
=N
0
×(
2
1
)
4
=
16
1
N
0
. The fraction decayed is 1−
16
1
=
16
16−1
=
16
15
.
Therefore, after 36 days,
16
15
of the atoms have decayed.
Final Answer: The final answer is
16
15
