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 are 11.1% more than the number of electrons) 

In the neutral of atom, number of electron. 

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 = . \(_{17}^{37}Cl\).