The dispersion (\(E(k)\)) of the conduction band (CB) and valence band (VB) for a semiconductor are shown schematically in the figure. Considering the possibility of an electron making a transition from the bottom of the CB to the top of the VB, which of the following options is/are correct?
Show Hint
The CB minimum and VB maximum sit at different values of \(k\) in the figure, so the transition needs a change in crystal momentum of \(\hbar q\).
A photon alone cannot supply that momentum change, so a phonon must be created alongside it, and the photon then carries less than \(E_g\).
A photon can be emitted with an energy exactly equal to \(E_g\).
A photon can be emitted with an energy less than \(E_g\).
A phonon can be created with a crystal momentum \(\hbar q\).
Show Solution
The Correct Option isC, D
Solution and Explanation
Step 1: Write down conservation laws for the transition.
Energy conservation requires \[ E_{CB}(k=q) - E_{VB}(k=0) = E_g \] to be released, and crystal momentum must change from $q$ to $0$, a change of $\hbar q$. Any emitted particles in the process must together balance both of these.
Step 2: Test a photon-only process.
If only a photon of energy $E_\gamma$ is emitted, energy conservation would need $E_\gamma = E_g$, while momentum conservation would need the photon momentum to equal $\hbar q$. A photon's momentum is $p = E_\gamma / c$, and for typical band gaps and lattice constants this is many orders of magnitude smaller than $\hbar q$. A photon-only process cannot satisfy both equations at once, so this pathway is ruled out, which is why statement (B), a photon energy of exactly $E_g$, fails as the actual process.
Step 3: Add a phonon to the balance.
Let a phonon of energy $E_{ph}$ and crystal momentum $\hbar q$ also take part. Momentum conservation is now satisfied exactly, since the phonon supplies the full $\hbar q$ the photon could not reach. Energy conservation becomes $E_\gamma + E_{ph} = E_g$, so $E_\gamma = E_g - E_{ph}$.
Step 4: Read off the answer for statement (C).
Since $E_{ph} > 0$ for a real phonon, $E_\gamma = E_g - E_{ph}$ works out strictly smaller than $E_g$. A photon can indeed be emitted with energy less than $E_g$, confirming statement (C) is TRUE, while statement (B) stays FALSE since the photon alone never carries the full $E_g$ in this process.
Step 5: Read off the answer for statement (D).
The phonon brought in during Step 3 to fix the momentum balance carries crystal momentum $\hbar q$ by construction. Since this is exactly what closes the momentum equation from Step 1, such a phonon can indeed be created, confirming statement (D) is TRUE.
Step 6: Read off the answer for statement (A).
Steps 3 and 4 show a consistent process, a photon plus a phonon together, that satisfies both conservation laws. So the transition is not forbidden outright, it only needs this two-particle route. Statement (A) is FALSE.
Final Answer:
Balancing energy and crystal momentum separately shows the transition needs a phonon to go with the photon, matching (C) and (D).
\[ \boxed{\text{C, D}} \]
