Question

The work function (ϕ) of a metal is the minimum energy needed to remove an electron from its surface. (a) Is it easier to remove an electron from a gaseous silver atom or from the surface of solid silver (ϕ = 7.59×10−19 J; IE = 731 kJ/mol)? (b) Explain the results in terms of the electron-sea model of metallic bonding.

Answer 1

Answer:

Explanation:

solid silver:

$7.59 e-19 J / silver atom \times (6.022e23 silver atom/ mol) \times ( 1kJ/1000 J) = 457 kJ/mol$

457 < 731; it is easier to remove electrons from surface of solid silver

b) overlap of orbitals results in a continuous energy band so that electrons can easily transition to higher energies and be removed.

Answer 2

The surface of solid silver is easier to extract electrons, and the further discussion can be defined as follows:

Working function:

$\to \phi = 7.59\times 10^{-19}\ J\\\\\to IE = 731 \ \frac{ kJ}{mol}$

For part a):

Calculating the solid silver:

$\to 7.59 e-{19} \ \frac{J }{ \text{silver atom}} \times (6.022 e^{23} \ \frac{\text{silver atom}}{mol}) \times ( 1\ \frac{kJ}{1000 \ J} ) \\\\\to 457\ \frac{kJ}{mol}\\\\\to 457 < 731$

Solid silver's surface makes it easier to extract electrons.

For part b):

The overlapping of orbitals creates a continuous energy range, permitting electrons to move into excitation frequencies and just be easily removed.

Find out more information about the function here:

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