Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A9 / a / iii

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- Solution iii

Solution

The energy-band widths are the difference between their extrema:

W_{s} = 4 t_{s} = 4∣ β_{s} ∣, W_{p} = 4 t_{p} = 4∣ β_{p} ∣.

For the separated bands, the top of the

s

band and bottom of the

p

band are both at

k = π / a

. Hence

E_{g}^{m i n} = α_{p} - α_{s} - 2 (t_{s} + t_{p}) .

It is a direct band gap at the zone boundary, because both limiting states have the same wavevector. This result uses the opposite signs of the

s

and axial-

p

hoppings.

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