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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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Verified Answer
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
There fore
or
Thus
For copper, EF = 700 eV
so that E = EF + ΔE = 7.00 eV - 0.19 eV = 6.81 eV
Most Upvoted Answer
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.
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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