Question

(a) Using the result of Problem 23 and 7.00eVfor copper’s Fermi energy, determine how much energy would be released by the conduction electrons in a copper coin with mass3.10g if we could suddenly turn off the Pauli exclusion principle. (b) For how long would this amount of energy light a100 Wlamp? (Note: There is no way to turn off the Pauli principle!)

Short answer

1. The amount of energy that would be released by the conduction electrons in a copper coin is $1.97\times {10}^4\mathrm{J}$.
2. This amount of energy will light a 100 lamp for 197 s.

Step-by-step solution

The given data

1. Fermi energy of copper, ${\mathrm{E}}_{\mathrm{F}}=7\mathrm{eV}$
2. Mass of copper coin,$\mathrm{m}=3.10\mathrm{g}$
3. Molar mass of copper,$\mathrm{A}=63.54\mathrm{g}/\mathrm{mol}$
4. Power or the rate of energy used of the lamp, P = 100 W

Understanding the concept of energy

The transfer of energy from one body to another in unit time interval is known as power. It is also described as the rate of doing work with respect to time.

Formulae:

The average energy used by the a body,${\mathrm{E}}_{\mathrm{avg}}=\frac{3}{5}{\mathrm{E}}_{\mathrm{F}}$ (i)

The total energy released by the conduction electrons,$\mathrm{E}={\mathrm{NE}}_{\mathrm{avg}}$ (ii)

where, N is the number of electrons present in the material and is the Fermi energy.

The rate of energy consumption, P = E /t (ii)

a) Calculation of the energy released by the conduction electrons

From equation (i) and equation (ii), we get -

$\mathrm{E}=\left(\frac{\mathrm{m}}{{\mathrm{AN}}_{\mathrm{A}}}\right)\left(\frac{3}{5}{\mathrm{E}}_{\mathrm{F}}\right)$

$$\begin{gathered} \left(\because \mathrm{where},N\mathit{=}\frac{\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathit{\left(}\mathit{m}\mathit{\right)}}{\mathit{m}\mathit{o}\mathit{l}\mathit{a}\mathit{r}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathit{\left(}\mathit{A}\mathit{\right)}\mathit{\times }\mathit{a}\mathit{v}\mathit{o}\mathit{g}\mathit{a}\mathit{d}\mathit{r}\mathit{o}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathit{\left(}{\mathit{N}}_{\mathit{A}}\mathit{\right)}}\right) \\ =\left(\frac{3.1\mathrm{g}}{\left(63.54g/mol\right)\left(6.022\times {10}^{23}/\mathrm{mol}\right)}\right)\left(\frac{3}{5}\left(7\mathrm{eV}\right)\left(1.6\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\right) \\ =1.97\times {10}^4\mathrm{J} \end{gathered}$$

Hence, the value of the energy is .

b) Calculation of the time till which the lamp is light on

Using the above value of energy, we can get the time till which the energy is lit up as follows:

$$\begin{gathered} t=\frac{1.97\times {10}^4\mathrm{J}}{100\mathrm{J}/\mathrm{s}} \\ =197\mathrm{s} \end{gathered}$$

Hence, the value of the time is 197 s .