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

Final Answer : Cl (-)

Steps and Understanding :
1) Let the no. of protons in ion (M-) be 'b'.
=> No. of electrons = b + 1
Let no. of neutrons be 'c'.

2) We have,
Mass No. = No. of protons + No of neutrons
=> 37 = b + c ------(1)

3) Then,
Ion has 11.1% more neutrons than electrons.
$= > c = (b + 1) + (b + 1) \frac{11.1}{100} \\ = > c = (b + 1) \times (1.111) - - - (2)$


4) Using, eq:(1) and (2) ,

$37 = b +1.111 (b + 1) \\ = > 37 = 2.111b +1 .111 \\ = > 35.889 = 2.111b \\ = > b = 17\: (approx)$
As, b is a natural number.


5) Then, we have
No. of electrons = b+1= 18
No. of neutrons = 37-17= 20
No. of protons = b = 17 (atomic number)

So, the symbol is Cl (-) .

Answer 2

$\fbox\textcolor{orange}{\texttt{Answer = Cl(37,17)}}$

$\textcolor{indigo}{\texttt{Explanation :-}}$

Suppose no. of electrons = x

no. of neutrons = $x + \frac{11.1}{100}\times{x}\: = \:1.111x$

No. of electrons in the neutral atom = $x - 1$

No. of protons in the neutral atom =$x - 1$

Mass no. = no. of neutrons + no. of protons

ie, $37\:=\:1.111x+x-1$

$2.111x\:=\:37+1\:=\:38$

$x = \frac{38}{2.111}\:=\texttt{18}$

No. of protons = Atomic no. =

$x-1\:=\:18-1\:=\texttt{17}$

Hence the symbol of the ion is: $</h3><h3>\fbox{^{37}_{17}Cl}$