In the relativistic free electron gas, the classical kinetic energy E = p²/2m is replaced by E = Vp?c² + m2c4 – mc?. If the electrons are ultra-relativistic, the rest mass contribution is ignored and the energy is simply written as E = pc. Repeat the calculation derived in class for the fermy energy of an ideal gas of free electrons, assuming the electrons are massless (ultra-relativistic).

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Q3.3 Please answer the following question throughly and detailed. Need to understand the concept.
In the relativistic free electron gas, the classical kinetic energy E = p² /2m is replaced by
E = Vp?c2 + m²c4 – mc². If the electrons are ultra-relativistic, the rest mass contribution is ignored and
the energy is simply written as E = pc. Repeat the calculation derived in class for the fermy energy of an
ideal gas of free electrons, assuming the electrons are massless (ultra-relativistic).
Transcribed Image Text:In the relativistic free electron gas, the classical kinetic energy E = p² /2m is replaced by E = Vp?c2 + m²c4 – mc². If the electrons are ultra-relativistic, the rest mass contribution is ignored and the energy is simply written as E = pc. Repeat the calculation derived in class for the fermy energy of an ideal gas of free electrons, assuming the electrons are massless (ultra-relativistic).
Expert Solution
Step 1

It is known that the kinetic energy for relativistic free electron is calculated as,

E=p2c2+m2c4-mc2

If we ignore the rest mass, m. Then, E=pc. Where p is the momentum and c is the speed of light.

From de-Broglie relation it is known that, p=ħk. Therefore, k=Eħc

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