Tardigrade
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Tardigrade
Question
Physics
The nuclei 613 C and 714 N can be described as
Q. The nuclei
6
13
C
and
7
14
N
can be described as
1736
188
Nuclei
Report Error
A
isotones
66%
B
isobars
13%
C
isomers
10%
D
isotopes
11%
Solution:
Number of neutrons for carbon
=
13
−
6
=
7
Number of neutrons for nitrogen
=
14
−
7
=
7
Both the nuclei have equal number of neutrons.
Hence, they are isotones.