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KCET 2015 Physics Solved Paper
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© examsnet.com
Question : 53
Total: 60
The half life of a radioactive substance is 20 minutes. The time taken between 50 % decay and 87.5 % decay of the substance will be
30 minutes
40 minutes
25 minutes
10 minutes
Validate
Solution:
Given, half-life of radioactive substance,
T
1
2
=
20
minutes
Using
N
=
N
0
e
−
λ
t
⇒
N
N
0
=
e
−
λ
t
⇒
−
λ
t
=
ln
(
N
N
0
)
⇒
t
=
−
1
λ
ln
(
N
N
0
)
We know
1
λ
=
T
1
∕
2
ln
2
t
=
−
ln
2
T
1
∕
2
ln
(
N
N
0
)
If
t
1
is time for 50% decay and
t
2
is time for 87.5% decay, we have
t
2
−
t
1
=
−
T
1
∕
2
ln
2
ln
(
N
2
N
0
)
+
T
1
∕
2
ln
2
ln
(
N
1
N
0
)
=
T
1
∕
2
ln
2
(
−
ln
N
2
+
ln
N
0
+
ln
N
1
−
ln
N
0
)
=
T
1
∕
2
ln
2
(
ln
N
1
−
ln
N
2
)
=
T
1
∕
2
ln
2
ln
(
N
1
N
2
)
Substituting the values, we get
t
2
−
t
1
=
20
m
i
n
0.693
ln
(
50
12.5
)
=
20
0.693
×
ln
4
=
40
minutes
Therefore, time taken between 50% decay and 87.5% decay will be 40 minutes.
© examsnet.com
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