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Question

The nuclei 13Al27 and 14Si28 are examples of:

A
isotopes
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B
isobars
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C
isotones
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D
isomers
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Solution

The correct option is C isotones
Number of neutrons present in 13Al27 =2713=14
Number of neutrons present in 14Si28 =2814=14
Since, elements which have equal number of neutrons are called as isotones, so 13Al27 and 14Si28 are isotones of each other.

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