Effective mass of electron in gold

The statement means that a conduction electron in gold responds to an applied force as if its mass were $m^* = 1.1\,m_e$ instead of the free-electron mass $m_e$. This is correct and is explained by the effective mass concept.

Reason

A conduction electron in a metal is not truly free. Besides the external force $F_{ext}$ (e.g. $-eE$), it feels the periodic force of the positive ion cores and other electrons, $F_{int}$:

$$F_{ext} + F_{int} = m_e a$$

Since $F_{int}$ cannot be written simply, its whole effect is put into a new mass:

$$F_{ext} = m^* a, \qquad m^* = \frac{\hbar^2}{d^2E/dk^2}$$

• For a perfectly free electron $E = \hbar^2k^2/2m_e$ and $m^* = m_e$.
• In gold, the E–k curve of the conduction band near the Fermi level is slightly less curved than the free-electron parabola, because of the interaction with the lattice. Smaller $d^2E/dk^2$ gives larger $m^*$: here $d^2E/dk^2 = \hbar^2/(1.1\,m_e)$, so $m^* = 1.1\,m_e$.

 E
 ^        free electron   gold band
 |        (m* = me)      (m* = 1.1 me,
 |          \   /         flatter)
 |           \ /   .  .
 |            V  .      .
 +---------------------------> k

Physical meaning

Under the same electric field, an electron in gold accelerates only $1/1.1 \approx 0.91$ times as fast as a free electron, as if it were 10 % heavier; the lattice "holds it back" slightly. The value is found from experiments such as electronic specific heat or cyclotron resonance. Since 1.1 is close to 1, the free electron model works quite well for gold (and other noble metals).