/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q12CQ The plot below shows the variati... [FREE SOLUTION] | 91Ó°ÊÓ

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The plot below shows the variation of ω with k for electrons in a simple crystal. Where, if anywhere, does the group velocity exceed the phase velocity? (Sketching straight lines from the origin may help.) The trend indicated by a dashed curve is parabolic, but it is interrupted by a curious discontinuity, known as a band gap (see Chapter 10), where there are no allowed frequencies/energies. It turns out that the second derivative ofω with respect to k is inversely proportional to the effective mass of the electron. Argue that in this crystal, the effective mass is the same for most values of k, but that it is different for some values and in one region in a very strange way.

Short Answer

Expert verified

The group velocity exceeds phase velocity in all regions except the regions near the jump.

Step by step solution

01

Definition of group and phase velocities

Group velocity is defined as the velocity of the whole envelope of a wave in space.

Phase velocity is defined as the velocity of a phase or part of a wave in space.

02

Explanation

Only within the regions near the jump does the slope of a line from the origin, ω/k, exceed the slope of the curve, or tangent line. Thus, only in these regions does the group velocity not exceed the phase velocity. Wherever the dispersion relation follows the parabolic plot, its second derivative is the same constant. Just for k values near the bandgap, it deviates.

03

Conclusion

For values of ω just above the gap, the second derivative is larger, and the effective mass is smaller. For ω values slightly below the gap, the second derivative is negative.

Therefore, the effective mass is negative.

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