Question:
For nonrelativistic electrons in a solid, different energy dispersion relations with effective masses \(m_a^*, m_b^*, m_c^*\) are schematically shown in the plots. Which one of the following options is CORRECT?
For nonrelativistic electrons in a solid, different energy dispersion relations with effective masses \(m_a^*, m_b^*, m_c^*\) are schematically shown in the plots. Which one of the following options is CORRECT?
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The effective mass of an electron is inversely proportional to the curvature of the energy dispersion relation. A steeper curve corresponds to a smaller effective mass.
Updated On: Aug 30, 2025
- \(m_a^* = m_b^* = m_c^*\)
- \(m_b^*>m_c^*>m_a^*\)
- \(m_c^*>m_b^*>m_a^*\)
- \(m_a^*>m_b^*>m_c^*\)
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The Correct Option is D
Solution and Explanation
- The energy dispersion relation \(\epsilon(k)\) for electrons in solids gives insight into the effective mass of the electrons. The curvature of the energy vs. wavevector plot (\(E(k)\)) is inversely related to the effective mass. The steeper the curvature, the smaller the effective mass.
- From the plots, we observe that the effective mass \(m_a^*\) corresponds to the steepest curvature, indicating the smallest effective mass, and \(m_c^*\) has the least curvature, indicating the largest effective mass. Thus, the correct order is \(m_a^*>m_b^*>m_c^*\).
- From the plots, we observe that the effective mass \(m_a^*\) corresponds to the steepest curvature, indicating the smallest effective mass, and \(m_c^*\) has the least curvature, indicating the largest effective mass. Thus, the correct order is \(m_a^*>m_b^*>m_c^*\).
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