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Question
An ion with mass number 37 possesses unit negative charge. If the ion contains 11.1% more neutrons than electrons. Find the symbol of the ion.
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Solution
Let the number of electrons in an ion = x
number of neutrons = n = x + `11.1/100` eV = 1.111 x
As the number of neutrons is 11.1% more than the number of electrons
In the neutral of the atom, a number of electrons.
e– = x – 1 (as the ion carries –1 charge)
Similarly number of protons = P = x – 1
Number of protons + number of neutrons = mass number = 37
(x – 1) + 1.111 x = 37
2.111 x = 37 + 1
2.111 x = 38
x = `38/2.111`
= 18.009
= 18
∴ Number of protons = atomic number – 1
= 18 – 1
= 17
∴ The symbol of the ion = \[\ce{^37_17Cl}\].
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