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Makovka662 [10]

3 years ago

12

Point y divides line XZ such that XY:YZ is 3:11. what fraction of XZ is XY?

1 answer:

Virty [35]3 years ago

7 0

$\bf X\stackrel{3}{\rule[0.35em]{7em}{0.25pt}}Y\stackrel{11}{\rule[0.35em]{23em}{0.25pt}}Z$

so the line looks more or less like that, with Y splitting XZ on a 3:11 ratio.

that simply means, that XZ is split on 3+11 pieces, and XY is taking up 3 of those pieces.

$\bf \cfrac{XZ}{XY}\implies \cfrac{3+11}{3}\implies \cfrac{14}{3}$

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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 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 $F(E_o) = 0.2689$

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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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