Respuesta :
Explanation:
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
Formula used for the radius of the [tex]n^{th}[/tex] orbit will be,
[tex]r_n=\frac{n^2\times 52.9}{Z}pm[/tex] (in pm)
where,
[tex]E_n[/tex] = energy of [tex]n^{th}[/tex] orbit
[tex]r_n[/tex] = radius of [tex]n^{th}[/tex] orbit
n = number of orbit
Z = atomic number
a) The Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
Z = 1
[tex]r_3=\frac{3^2\times 52.9}{1} pm[/tex]
[tex]r_3=476.1 pm[/tex]
476.1 pm is the Bohr radius of an electron in the n = 3 orbit of a hydrogen atom.
b) The energy (in J) of the atom in part (a)
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
[tex]E_3=-13.6\times \frac{1^2}{3^2}eV=1.51 eV[/tex]
[tex]1 eV=1.60218\times 10^{-19} Joules[/tex]
[tex]1.51 eV=1.51\times 1.60218\times 10^{-19} Joules=2.4210\times 10^{-19} Joules[/tex]
[tex]2.4210\times 10^{-19} Joules[/tex] is the energy of n = 3 orbit of a hydrogen atom.
c) The energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
n = 3, Z = 3
[tex]E_3=-13.6\times \frac{3^2}{3^2}eV = -13.6 eV[/tex]
[tex]=-13.6eV = -13.6\times 1.60218\times 10^{-19} Joules=2.179\times 10^{-18} Joules[/tex]
[tex]2.179\times 10^{-18} Joules[/tex] is the energy of an Li²⁺ ion when its electron is in the n = 3 orbit.
d) The difference in answers is due to change value of Z in the formula which is am atomic number of the element..
[tex]E_n=-13.6\times \frac{Z^2}{n^2}eV[/tex]
