Question

How do you calculate the wavelength of light (in nm) required for mercury to emit an electron?

Answer 1

You must look up the ionization energy for mercury and use Planck's relation `E=(hc)/lambda` to find the wavelength that corresponds to this energy.
The result will be `1.18xx10^(-7) m` (this is ultraviolet light)

Explanation:

Since atoms can only absorb energy in discrete quantities, called photons, the energy of a single photon must be sufficient to provide the ionization energy of the atom. If not, the electron will jump into a higher bound state (a higher energy orbital) but will soon return to the ground state, with emission of a photon.

So, we require the ionization energy of mercury to do this calculation. You look this up - it is 1007 kJ/mol.

So, the minimum energy of a photon that will be able to remove an electron from mercury is this same value, 1007 kJ/mol.

We place this into Planck's relation for the energy and wavelength of absorbed photons

`E=(hc)/lambda`

Since the units on ionization energy are kJ per mol, we must convert Planck's constant into a kJ s per mol value.

`h=(6.63xx10^(-34) Js)*((6.02xx10^(23)) /(mol))-:(1000 J/(kJ))=3.99xx10^(-13)`

`1007=((3.99xx10^(-13))(3xx10^8))/lambda`

`lambda=((3.99xx10^(-13))(3xx10^8))/1007=1.18xx10^(-7) m`