Q.

Consider an imaginary ion  2248X3. The nucleus contains ‘a’% more neutrons than the number of electrons in the ion. The value of ‘a’ is ____. [nearest integer]

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answer is 4.

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Detailed Solution

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 2248X3

No. of neutrons = 26

No. of electrons = 25

% of extra neutrons=262525×100=4

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