Question

Ionization energy is the energy required to remove an electron from an atom in the gas phase. The ionization energy of gold is 890.1 kJ/mol. Calculate the maximum wavelength of light that will ionize gold.

Answer 1

Ionization of Gold:

The maximum wavelength of light that will ionize gold is 134.5 x 10 ⁻⁴m  

Calculation:

1 mole requires = 890.1kJ = 8.901×10⁵ J

The energy needed to ionize a single atom = Divide it by Avogadro's number (6.0221409×10²³)

Thus, 1 atom requires = 8.901×10⁵ J / 6.0221409×10²³ J

                                    =  1.478×10⁻¹⁸ J

The Energy of photon = E = hc/λ   where,

h =  Planck constant (6.626070040Ă—10â’34 Js)

λ =  the wavelength

c = the speed of light

Putting the values:

E = 1.478×10⁻¹⁸ J

h =6.63x10⁻³⁴ Js

c = 3 x 10⁸ m/s

λ = ?

λ = hc/E

  = (6.63x10⁻³⁴ x 3 x 10⁸ ) / 1.478×10⁻¹⁸

  = 134.5 x 10 ⁻⁴m  

 

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