Consider an imaginary ion {}_{22}^{48}X^{3-}. | vedclass

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Question

(A) For the ion ${}_{22}^{48}X^{3-}$:Atomic number $(Z)$ = $22$.Mass number $(A)$ = $48$.Number of neutrons = $A - Z = 48 - 22 = 26$.Number of electro

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) For the ion ${}_{22}^{48}X^{3-}$:Atomic number $(Z)$ = $22$.Mass number $(A)$ = $48$.Number of neutrons = $A - Z = 48 - 22 = 26$.Number of electrons = $Z + 3 = 22 + 3 = 25$.We are given that the nucleus contains '$a$'$\%$ more neutrons than the number of electrons.Percentage difference = $\frac{	ext{Number of neutrons} - 	ext{Number of electrons}}{	ext{Number of electrons}} 	imes 100$.Percentage difference = $\frac{26 - 25}{25} 	imes 100 = \frac{1}{25} 	imes 100 = 4$.Thus,the value of '$a$' is $4$.

Consider an imaginary ion ${}_{22}^{48}X^{3-}$. 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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