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