Question

Calculate the Fermi energy of Cu at 0K if the concentration of electron is 8.5×1028 m–3

Answer 1

Answer:

So the energy given to an electron by the electric field by 100 volts applied to a 1 meter copper wire would be on the order of W=eEd = 100 volts x 40 nm = 0.000004 eV.

Explanation:

Try to solve using this way

Answer 2

Given:

concentration (n) of the Cu = 8.5 × 10²⁸ ${m}^{ - 3}$

To Find:

The Fermi energy of Cu at 0K.

Solution:

• Fermi energy refers to the maximum kinetic energy that can be attained by the atoms at 0K.

• It can be mathematically represented as:

Ef = $\frac{ {h}^{2} }{8m} {( \frac{3n}{\pi} )}^{ \frac{2}{3} }$

where,

h = 6.63 × 10^-34 Js (Planck's constant)

mass of electron = 9.11 × 10^-31 kg

n (concentration) = 8.5 × 10²⁸ m^-3

• Put the above values in the formula of Ef to find the Fermi energy

$= \frac{ {h}^{2} }{8m} {( \frac{3n}{\pi} )}^{ \frac{2}{3} }$

$= \frac{ {6.63 \times {10}^{ - 34} }^{2} }{8 \times 9.11 \times {10}^{ - 31} } {( \frac{3(8.5 \times {10}^{28} }{\pi} )}^{ \frac{2}{3} }$

$= 1.1307 \times {10}^{ - 18} j$

• Now convert this to eV

1 joule = 6.242 × 10¹⁸ Electron volts

1.1307 × 10^-18 joule = 1.1307 × 10^-18 × 6.242 × 10¹⁸ eV

                                 = 7.0578 eV

Hence, the Fermi Energy of Cu at 0k is 7.0578 eV.

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