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: Understanding the Concept:
In a solid, an electron's response to forces is set by the curvature of its energy band \(E(k)\), not its free space mass. This is captured by the effective mass, defined from how fast the group velocity changes with \(k\).

Step 2: Key Formula or Approach:
The effective mass formula is
\[ \frac{1}{m^*} = \frac{1}{\hbar^2}\frac{d^2E}{dk^2} \]
So the second derivative of \(E(k) = E_0 - 2\gamma\cos(ka)\) with respect to \(k\) is needed first.

Step 3: Detailed Explanation:
Differentiate once:
\[ \frac{dE}{dk} = 2\gamma a \sin(ka) \]
Differentiate again:
\[ \frac{d^2E}{dk^2} = 2\gamma a^2 \cos(ka) \]
At \(k = 0\), \(\cos(0) = 1\), so:
\[ \frac{d^2E}{dk^2}\bigg|_{k=0} = 2\gamma a^2 \]
This gives the effective mass:
\[ m^* = \frac{\hbar^2}{2\gamma a^2} \]
Now put in the numbers, with \(\gamma = 0.5\) eV \(= 0.5 \times 1.6 \times 10^{-19} = 8 \times 10^{-20}\) J and \(a = 0.5\) nm \(= 5 \times 10^{-10}\) m, so \(a^2 = 2.5 \times 10^{-19}\) m\(^2\):
\[ m^* = \frac{(1.054 \times 10^{-34})^2}{2 \times (8 \times 10^{-20}) \times (2.5 \times 10^{-19})} = \frac{1.111 \times 10^{-68}}{4 \times 10^{-38}} = 2.777 \times 10^{-31} \text{ kg} \]
The ratio to the free electron mass is:
\[ R_0 = \frac{m^*}{m_e} = \frac{2.777 \times 10^{-31}}{9.1 \times 10^{-31}} = 0.305 \]

Final Answer:
Rounded to two decimal places, the effective mass at the bottom of this band comes out to about 0.31 times the free electron mass. \[ \boxed{R_0 = 0.31} \]