Respuesta :
Answer:
[tex]\boxed{8.4\times 10^{-10} \text{ N}}[/tex]
Explanation:
The formula for the force exerted between two ions is
[tex]F = \dfrac{kq_{1}q_{2}}{r^{2}} = \dfrac{kz_{1}z_{2}e^{2}}{r^{2}}[/tex]
where the Coulomb constant
k = 8.988 × 10⁹ N·m·C⁻²
Data:
z₁ = +2; r₁ = 0.62 nm
z₂ = -1; r₂ = 0.121 nm
Calculations:
r = r₁ + r₂ = 0.62 + 0.121 = 0.741 nm
[tex]F =\dfrac{8.988 \times 10^{9} \text{ N $\cdot$ m$^{2} \cdot$ C$^{-2}$} \times 2 \times (-1) \times ( 1.602 \times 10^{-19} \text{ C})^{2}}{(0.741 \times 10^{-9} \text{ m})^{2}}\\\\ = -\mathbf{8.4\times 10^{-10}} \textbf{ N}\\\text{The attractive force between the ions is } \boxed{\mathbf{8.4\times 10^{-10}} \textbf{ N}}[/tex]
The force of attraction between these two ions at their equilibrium interionic separation is [tex]-8.39\times 10^{-10} N[/tex].
Explanation:
Given:
The divalent cation and monovalent anion with atomic radii of 0.62 nm and 0.121 nm, respectively.
To find:
Force of attraction between two ions at their equilibrium interionic separation.
Solution:
The charge on divalent cation = [tex]z_1= +2[/tex]
The charge on monovalent anion = [tex]z_2= -1[/tex]
Atomic radius of divalent cation = [tex]r_c=0.62 nm[/tex]
Atomic radius of monovalent anion = [tex]r_a=0.121nm[/tex]
Charge on an electron = [tex]e = -1.6\times 10^{-19} C[/tex]
The inter ionic seperation between ions = d
[tex]d=r_c+r_a\\d=0.62 nm + 0.121 nm = 0.741 nm\\1 nm=10^{-9}m\\0.741 nm=0.741\times 10^{-9}m\\=7.41\times 10^{-10} m[/tex]
The permittivity of vacuum =[tex]\epsilon_o=8.85\times 10^{-12} C^2/Nm^2[/tex]
The force of attraction between ions is given by:
[tex]F=\frac{z_1z_2e^2}{4\pi \epsilon _o d^2}\\F=\frac{(+2)(-1)\times ( -1.6\times 10^{-19} C)^2}{4\times 3.14\times 8.85\times 10^{-12} C^2/Nm^2\times (7.41\times 10^{-10} m)^2}\\F=-8.39\times 10^{-10} N[/tex]
(negative sign indicates attractive force)
The force of attraction between these two ions at their equilibrium interionic separation is [tex]-8.39\times 10^{-10} N[/tex].
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