Question

The ionization energy for potassium 419 KJ/mol. The wavelength of light emitted when an excited K atom undergoes the 4s<---- 4p transition is approximately 769 nm. Using this info, calculate the energies of the 4s and 4p orbitals in potassium.

Answer 1

Answer : The energies of the 4s and 4p orbitals in potassium is,
`-6.96* 10^(-19)J\text{ and }-4.37* 10^(-19)J`

Solution :

First we have to calculate the change in energy.

Formula used :

`\Delta E=(h* c)/(\lambda)`

where,

`\Delta E` = change in energy

h = Planck's constant =
`6.626* 10^(-34)Js`

c = speed of light =
`3* 10^(8)ms^(-1)`

`\lambda` =
`769nm=769* 10^(-9)m`

Now put all the given values in this formula, we get

`E=((6.626* 10^(-34)Js)* (3* 10^(8)ms^(-1)))/(769* 10^(-9)m)=2.585* 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}=(419* 10^3J/mol)/(6.022* 10^(23))=6.958* 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* 10^(-19)J`

Now we have to calculate the energy of the 4p orbital in potassium.

`\Delta E=E_(4p)-E_(4s)`

`2.585* 10^(-19)J=E_(4p)-(-6.958* 10^(-19)J)`

`E_(4p)=-4.37* 10^(-19)J`

Therefore, the energies of the 4s and 4p orbitals in potassium is,
`-6.96* 10^(-19)J\text{ and }-4.37* 10^(-19)J`