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.

Answer 1

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 = \[\ce{^37_17Cl}\].