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3 years ago

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A plumber charges a flat fee for each job, plus an hourly rate for the number of hours the job takes to complete. The total cost

of the job, in dollars can be modeled by the expression 50 + 65x. What does the constant term in the expression represent in this situation?

1 answer:

Burka [1]3 years ago

3 0

Answer:

X=number of hours

Step-by-step explanation:

$50 is the flat fee

$65 per hour

X how many hours it took for the job.

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evaluate the fermi function for an energy KT above the fermi energy. find the temperature at which there is a 1% probability tha

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Complete Question

Evaluate the Fermi function for an energy kT above the Fermi energy. Find the temperature at which there is a 1% probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron.

Answer:

a

The Fermi function for the energy KT is $F(E_o) = 0.2689$

b

The temperature is $T_k = 1261 \ K$

Step-by-step explanation:

From the question we are told that

The energy considered is $E = 0.5 eV$

Generally the Fermi function is mathematically represented as

$F(E_o) = \frac{1}{e^{\frac{[E_o - E_F]}{KT} } + 1 }$

Here K is the Boltzmann constant with value $k = 1.380649 *10^{-23} J/K$

$E_F$ is the Fermi energy

$E_o$ is the initial energy level which is mathematically represented as

$E_o = E_F + KT$

So

$F(E_o) = \frac{1}{e^{\frac{[[E_F + KT] - E_F]}{KT} } + 1}$

=> $F(E_o) = \frac{1}{e^{\frac{KT}{KT} } + 1}$

=> $F(E_o) = \frac{1}{e^{ 1 } + 1}$

=> $F(E_o) = 0.2689$

Generally the probability that a state, with an energy 0.5 eV above the Fermi energy, will be occupied by an electron is mathematically represented by the Fermi function as

$F(E_k) = \frac{1}{e^{\frac{[E_k - E_F]}{KT_k} } + 1 } = 0.01$

Here $E_k$ is that energy level that is 0.5 ev above the Fermi energy $E_k = 0.5 eV + E_F$

=> $F(E_k) = \frac{1}{e^{\frac{[[0.50 eV + E_F] - E_F]}{KT_k} } + 1 } = 0.01$

=> $\frac{1}{e^{\frac{0.50 eV ]}{KT_k} } + 1 } = 0.01$

=> $1 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} } + 0.01$

=> $0.99 = 0.01 * e^{\frac{0.50 eV ]}{KT_k} }$

=> $e^{\frac{0.50 eV ]}{KT_k} } = 99$

Taking natural log of both sides

=> $\frac{0.50 eV }{KT_k} } =4.5951$

=> $0.50 eV =4.5951 * K * T_k$

Note eV is electron volt and the equivalence in Joule is $eV = 1.60 *10^{-19} \ J$

So

$0.50 * 1.60 *10^{-19 } =4.5951 * 1.380649 *10^{-23} * T_k$

=> $T_k = 1261 \ K$

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3 years ago