Respuesta :
Answer:
B. [tex]2.18\times 10^{-18}\ J[/tex]
Explanation:
Ionization energy is the minimum amount of energy which is required to knock out the loosely bound valence electron from the isolated gaseous atom.
The expression for the energy of an electron in the nth orbit is:-
[tex]E_n=-2.18\times 10^{-18}\times \frac{1}{n^2}\ Joules[/tex]
For transitions:
[tex]Energy\ Difference,\ \Delta E= E_f-E_i =-2.18\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
[tex]\Delta E=2.18\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J[/tex]
Given, [tex]n_i=1\ and\ n_f=inifinity[/tex]
[tex]\Delta E=2.18\times 10^{-18}(\frac{1}{1^2} - \dfrac{1}{\infty^2})\ J[/tex]
[tex]\Delta E=2.18\times 10^{-18}\ J[/tex]
Ionization energy, [tex]\Delta E=2.18\times 10^{-18}\ J[/tex]
