Atoms of $_{7} A,_{8} B a n d_{9} C$ are such that $_{8} B$ is an isobar of $_{7} A$ and atom $_{9}^{17} C$ is isotone to $_{8} B$ . The no. of neutrons in $A$ are:

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Solution

Isotones contain same number of neutrons and different number of protons.
$_{9}^{17} C$ and $_{8} B$ are isotones.
They contain $17 - 9 = 8$ neutrons.
Hence, mass number of B is $8 + 8 = 16$
Isobars have same mass number but different atomic number.
$_{8}^{16} B$ and $_{7} A$ are isobars.
The mass number of A will be same as that of B i.e, 16.
The number of neutrons in A $= 16 - 7 = 9$

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Atoms of 7A,8Band9C are such that 8B is an isobar of 7A and atom 179C is isotone to 8B. The no. of neutrons in A are:

Atoms of $_{7} A,_{8} B a n d_{9} C$ are such that $_{8} B$ is an isobar of $_{7} A$ and atom $_{9}^{17} C$ is isotone to $_{8} B$ . The no. of neutrons in $A$ are: