Answer : The energies of the 4s and 4p orbitals in potassium is, [tex]-6.96\times 10^{-19}J\text{ and }-4.37\times 10^{-19}J[/tex]
Solution :
First we have to calculate the change in energy.
Formula used :
[tex]\Delta E=\frac{h\times c}{\lambda}[/tex]
where,
[tex]\Delta E[/tex] = change in energy
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^{8}ms^{-1}[/tex]
[tex]\lambda[/tex] = [tex]769nm=769\times 10^{-9}m[/tex]
Now put all the given values in this formula, we get
[tex]E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^{8}ms^{-1})}{769\times 10^{-9}m}=2.585\times 10^{-19}J[/tex]
The difference in energy levels is defined as,
[tex]\Delta E=E_{4p}-E_{4s}[/tex]
Now we have to calculate the first ionization energy for one atom of potassium.
[tex]I.E=\frac{\text{Ionization energy for potassium}}{Avogadro's number}=\frac{419\times 10^3J/mol}{6.022\times 10^{23}}=6.958\times 10^{-19}J[/tex]
Now this ionization energy will be equal to the negative of the orbital energy of the electron located in the 4s-orbital.
[tex]E_{4s}=-I.E=-6.958\times 10^{-19}J[/tex]
Now we have to calculate the energy of the 4p orbital in potassium.
[tex]\Delta E=E_{4p}-E_{4s}[/tex]
[tex]2.585\times 10^{-19}J=E_{4p}-(-6.958\times 10^{-19}J)[/tex]
[tex]E_{4p}=-4.37\times 10^{-19}J[/tex]
Therefore, the energies of the 4s and 4p orbitals in potassium is, [tex]-6.96\times 10^{-19}J\text{ and }-4.37\times 10^{-19}J[/tex]
