Assertion–Reason (Fermi–Dirac statistics): At any nonzero temperature, the probability that an energy state E equal to the Fermi level E_F is occupied equals 0.5. Reason: The Fermi–Dirac function f(E) = 1 / (1 + exp[(E − E_F) / (kT)]) gives f(E_F) =.

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:The Fermi–Dirac distribution governs electron occupancy in solids. Understanding occupancy at the Fermi level is central to band theory, density of states calculations, and semiconductor device modeling. Given Data / Assumptions: Temperature T > 0 K. Fermi level E_F defined such that f(E_F) = 0.5 at any finite T. Boltzmann constant k and the standard Fermi–Dirac form are used. Concept / Approach:The Fermi–Dirac probability is f(E) = 1 / (1 + exp[(E − E_F)/(kT)]). Substituting E = E_F makes the exponent zero, giving f = 1/2. This result holds for all T > 0 K. At T = 0 K, all states with E < E_F are fully occupied (f = 1) and those with E > E_F are empty (f = 0), while E = E_F is a boundary case. Step-by-Step Solution: Write f(E) = 1 / (1 + exp[(E − E_F)/(kT)]).Set E = E_F ⇒ exponent = 0.Compute f(E_F) = 1 / (1 + 1) = 0.5.Therefore the assertion is true and the reason directly explains it. Verification / Alternative check: Plot f(E) vs. E around E_F for different T; all curves cross at (E_F, 0.5). Why Other Options Are Wrong: Any statement denying f(E_F) = 0.5 at finite T contradicts the Fermi–Dirac formula. Common Pitfalls: Confusing T = 0 K step function with finite-temperature smearing; forgetting to evaluate the exponential at zero. Final Answer: Both A and R are true and R is correct explanation of A

Assertion–Reason (Fermi–Dirac statistics): At any nonzero temperature, the probability that an energy state E equal to the Fermi level E_F is occupied equals 0.5. Reason: The Fermi–Dirac function f(E) = 1 / (1 + exp[(E − E_F) / (kT)]) gives f(E_F) = 1 / (1 + exp(0)) = 0.5.

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