Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A9 / b / iii

Table of contents
- Solution iii

Solution

Substitution in part (a) gives

k_{F} = 2 π (4.77 \times 1 0^{19}) ≃ 1.73 \times 1 0^{10} m^{- 1},

and

E_{F} = \frac{π ℏ ^{2} ( 4.77 \times 1 0 ^{19} )}{m _{e}} ≃ 1.83 \times 1 0^{- 18} J ≃ 11.4 eV .

The conversion uses

1 eV = 1.60218 \times 1 0^{- 19} J

. These are the predictions of the stipulated two-dimensional free electron model, with the vacuum electron mass and one carrier per inferred square cell.

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