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For copper at 1000 K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (EF = 7 eV)
    Correct answer is '6.81'. Can you explain this answer?
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    For copper at 1000K, the energy will be _____ eV at which the probabil...
    The probability F(E) of a state corresponding to energy E being occupied by an electron temperature T is given by

    There fore 
    or 
    Thus

    For copper, EF = 700 eV
    so that E = EF + ΔE = 7.00 eV - 0.19 eV = 6.81 eV
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    For copper at 1000K, the energy will be _____ eV at which the probabil...
    Calculating the Energy at which the Probability is 0.90 for Copper at 1000K

    Given information:
    - Temperature (T) = 1000K
    - Fermi energy (EF) = 7 eV
    - Probability (F(E)) = 0.90

    To find the energy at which the probability F(E) is 0.90, we can use the Fermi-Dirac distribution function:

    F(E) = 1 / (1 + exp((E - EF) / kT))

    where:
    - E is the energy
    - EF is the Fermi energy
    - k is the Boltzmann constant (8.6173 × 10^-5 eV/K)
    - T is the temperature

    Now, we need to solve the equation F(E) = 0.90 for E. Rearranging the equation, we get:

    0.90 = 1 / (1 + exp((E - EF) / kT))

    Multiplying both sides by (1 + exp((E - EF) / kT)), we have:

    0.90 * (1 + exp((E - EF) / kT)) = 1

    Dividing both sides by 0.90, we get:

    1 + exp((E - EF) / kT) = 1 / 0.90

    Subtracting 1 from both sides, we have:

    exp((E - EF) / kT) = 1 / 0.90 - 1

    Taking the natural logarithm (ln) of both sides, we get:

    (E - EF) / kT = ln(1 / 0.90 - 1)

    Multiplying both sides by kT, we have:

    E - EF = kT * ln(1 / 0.90 - 1)

    Finally, adding EF to both sides, we get the expression for E:

    E = EF + kT * ln(1 / 0.90 - 1)

    Now, let's substitute the given values into the equation:

    E = 7 eV + (8.6173 × 10^-5 eV/K * 1000K * ln(1 / 0.90 - 1))

    Calculating the right side of the equation:

    E ≈ 6.81 eV

    Therefore, at 1000K, the energy at which the probability F(E) is 0.90 for copper is approximately 6.81 eV.
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    For copper at 1000K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (EF = 7 eV)Correct answer is '6.81'. Can you explain this answer?
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    For copper at 1000K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (EF = 7 eV)Correct answer is '6.81'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about For copper at 1000K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (EF = 7 eV)Correct answer is '6.81'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For copper at 1000K, the energy will be _____ eV at which the probability F(E) that a conduction electron state will be occupied is 0.90. (EF = 7 eV)Correct answer is '6.81'. Can you explain this answer?.
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