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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ABoth A and R are true and R is correct explanation of A
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BBoth A and R are true but R is not correct explanation of A
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CA is true but R is false
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DA is false but R is true
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EBoth A and R are false
Answer
Correct Answer: Both A and R are true and R is correct explanation of A
Explanation
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