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Nuclei
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First Order Radioactive Decay
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The activity of a nuclide is 15 millicur
Question
The activity of a nuclide is
1
5
millicurie. If its decay constant is
0
.
0
0
5
sec
−
1
, the number of atoms present in it is
A
11.1 x 10
1
0
B
1 x 10
9
C
111 x 10
1
0
D
1.1 x 10
1
0
Medium
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Updated on : 2022-09-05
Solution
Verified by Toppr
Correct option is A)
Decay constant
=
0
.
0
0
5
s
e
c
−
1
=
λ
Activity of nuclei
=
1
5
×
3
.
7
×
1
0
7
d
p
s
[
∵
1
m
i
l
l
i
c
u
r
i
e
=
3
.
7
×
1
0
7
d
p
s
]
=
5
.
5
5
×
1
0
8
d
p
s
Now, from differential rate law:
d
t
d
N
=
λ
N
[ where N is no of atoms present]
⇒
0
.
0
0
5
5
.
5
5
×
1
0
8
=
N
⇒
N
=
1
1
.
1
×
1
0
1
0
a
t
o
m
s
Video Explanation
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