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2 years ago

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A bank manager wants to encourage new customers to open accounts with initial deposits of at least $3000. He has posters made fo

r the promotion. Under the new promotion, what is the minimum amount of interest a new account would make in one year it there were no withdrawals from the account?

1 answer:

belka [17]2 years ago

7 0

Incomplete question. However, let us assume

the interest rate is 5%, and
the time is 1 year.

Answer:

<u>$150</u>

Step-by-step explanation:

Note, the bank manager wants new customer accounts opened with at least, $3,000. Hence we have a principal, $3000. Using the simple interest formula:

$S.I= \frac{amount*rate*time}{100}$

$SI=\frac{3000*5*1}{100}$ = $150

Therefore, the <em>minimum</em> amount of interest a new account would make in one year is <u>$150.</u>

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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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