The ionization energy is 111.15 [tex]*10^{-18}[/tex] eV.
Explanation:
Ionization energy is the measure of amount of energy required to remove an electron from valence shell. So depending upon the shell and the number of electrons to be removed , the ionization energy will vary. Thus, increase in atomic number will lead to decrease in ionization energy as the increase in number of shells will lead to decrease in the attractive force between nucleus and the shell leading to decrease in ionization energy.
Using Bohr atom model,
[tex]radius = \frac{0.59n^{2} }{Z}[/tex]
Here n is the valence shell principle quantum number and Z is the atomic number.
As the radius is given as 0.23 nm, we can determine the ratio of n²/Z. This value can be substituted in the formula for ionization energy.
Ionization energy =[tex]\frac{-2.178*10^{-18}*Z^{2} }{n^{2} }[/tex]
As radius = 0.23 nm, the Z/n² value is
[tex]0.23=\frac{0.59*n^{2} }{Z}[/tex]
[tex]\frac{Z}{n^{2} }=\frac{0.59}{0.23} =2.56[/tex]
As, the atomic number of Ca is 20, then
I.E = [tex]-2.178*10^{-18}*2.56*20 = 111.15 * 10^{-18}[/tex]
So the ionization energy is 111.15 [tex]*10^{-18}[/tex] eV.
The magnitude of ionization energy be "[tex]111.15\times 10^{-18}[/tex] eV".
According to the question,
Radius of calcium atom = 0.23 nm
By using Bohr atom model, we get
→ Radius = [tex]\frac{0.59n^2}{Z}[/tex]
By substituting the values, we get
[tex]0.23 = \frac{0.59\times n^2}{Z}[/tex]
[tex]\frac{Z}{n^2} = \frac{0.59}{0.23} = 2.56[/tex]
We know the atomic number of Ca = 20
→ Ionization energy = [tex]-2.178\times 10^{-18}\times 2.56\times 20[/tex]
= [tex]111.15\times 10^{-8}[/tex] eV
Thus the above answer is correct.
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