Question

For the energy dispersion of an electron in a one-dimensional solid \(E(k) = E_0 - 2\gamma \cos(ka)\), the ratio of the effective mass of the electron in the solid to the free electron mass (\(m_e\)) at \(k = 0\) is \(R_0\). Taking \(\gamma = 0.5\) eV and \(a = 0.5\) nm, the value of \(R_0\) (rounded off to two decimal places) is
(\(\hbar = 1.054 \times 10^{-34}\) J.s, \(m_e = 9.1 \times 10^{-31}\) kg, electron charge \(= 1.6 \times 10^{-19}\) C)

Hint:

Hint:
The effective mass comes from the curvature of the band, \(m^* = \hbar^2 / (d^2E/dk^2)\), evaluated at \(k=0\).

Answer 1

Correct Answer: 0.31

Step 1: Replace the cosine band with its parabola near $k=0$.
Near the bottom of the band $k$ is small, so use the small angle expansion $\cos(ka) \approx 1 - \dfrac{(ka)^2}{2}$. Putting this into $E(k) = E_0 - 2\gamma\cos(ka)$ gives
\[ E(k) \approx E_0 - 2\gamma + \gamma a^2 k^2 \]
This is a plain parabola in $k$, just like a free electron, but with a different number in front of $k^2$.

Step 2: Match the parabola to the free electron form.
A free electron has $E(k) = \dfrac{\hbar^2 k^2}{2m}$, so matching the coefficient of $k^2$ in both expressions gives
\[ \gamma a^2 = \frac{\hbar^2}{2m^*} \quad \Rightarrow \quad m^* = \frac{\hbar^2}{2\gamma a^2} \]
This is the same expression the curvature formula gives, but reached by comparing the shape of the band instead of differentiating twice.

Step 3: Put in the numbers.
With $\gamma = 0.5$ eV $= 8 \times 10^{-20}$ J and $a = 0.5$ nm $= 5 \times 10^{-10}$ m (so $a^2 = 2.5 \times 10^{-19}$ m$^2$):
\[ m^* = \frac{(1.054 \times 10^{-34})^2}{2 (8 \times 10^{-20})(2.5 \times 10^{-19})} = 2.777 \times 10^{-31} \text{ kg} \]

Step 4: Take the ratio.
\[ R_0 = \frac{m^*}{m_e} = \frac{2.777 \times 10^{-31}}{9.1 \times 10^{-31}} = 0.305 \]

Final Answer:
Rounded off, the ratio comes to 0.31, the same value the curvature formula gives. \[ \boxed{R_0 = 0.31} \]