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VITEEE 2014 Paper
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Section:
Physics
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Question : 28
Total: 120
The activity of a radioactive sample is measured as
N
0
counts per minute at t = 0 and
N
0
/C counts per minute at t = 5 min. The time, (in minute) at which the activity reduces to half its value, is
l
o
g
e
2
5
5
l
o
g
e
2
5
l
o
g
10
2
5
l
o
g
e
2
Validate
Solution:
After n half-lives
N
N
0
=
(
1
2
)
n
=
(
1
2
)
t
â
7
N =
N
+
0
e
=
c
N
0
c
N
0
=
(
1
2
)
5
â
7
â
1
e
=
(
1
2
)
5
â
7
Taking log on both sides, we get
log 1 - log e =
5
7
log
1
2
- 1 =
5
7
(- log 2)
T =
5
l
o
g
e
2
Now, let tâ be the time after which activity reduces to half
(
1
2
)
=
(
1
2
)
1
â²
â
5
l
o
g
e
2
t' = 5
l
o
g
e
2
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