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Tardigrade
Question
Physics
The nuclei 6 C 13 and 7 N 14 can be described as
Q. The nuclei
6
C
13
and
7
N
14
can be described as
3018
232
AIPMT
AIPMT 1990
Atoms
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A
isotones.
85%
B
isobars
7%
C
isotopes of carbon
5%
D
isotopes of nitrogen
3%
Solution:
As
6
C
13
and
7
N
14
have same no. of neutrons
(
13
−
6
=
7
for
C
and
14
−
7
=
7
for
N
)
, they are isotones.