Question

derive an expression of an effective mass of an electron ​

Answer 1

Answer:

This mass of an electron is called the effective mass of an electron, denoted as m*. The effective mass is thus determined by d 2E/dK 2. From band theory of solids, we know E is not proportional to K 2.

Answer 2

Answer:

The effective mass is thus determined by $d^2E/dK^2$

Explanation:

Step 1: Based on a free electron model, the density of states and electron transit are conveniently described by the effective mass, a handy descriptor of the electronic band structure.

Step 2: The effective mass, which is different from the free electron mass and refers to the mass of an electron in the periodic potentials of a crystal, is used to describe it. The de Broglie hypothesis states that an electron in motion is connected to a wave. An electron's (v) velocity is the same as its associated wave's (vg) group velocity. The group velocity is calculated using:

$v_g=\frac{d \omega}{d k}$ .....(i)

where ω is the angular frequency (2πν) and K is the propagation vector of the wave.

In quantum mechanics, the energy, ‘E’ of an electron is given by:

$E=\hbar \omega$ ......(ii)

Substituting value of w from eq (2) to eq (1), we get

$v_g=\frac{1}{\hbar} \frac{d E}{d \hbar}$........(iii)

Differentiating Equation (3)  with respect to ‘t’ , we get acceleration of electron as

$\begin{aligned}& a=\frac{\mathrm{d} v_g}{\mathrm{~d} t}=\frac{1}{\hbar} \frac{\mathrm{d}}{\mathrm{d} t}\left(\frac{d E}{d k}\right) \\& a=\frac{1}{\hbar} \frac{\mathrm{d}}{\mathrm{d} k}\left(\frac{d E}{d k}\right) \frac{\mathrm{dk}}{\mathrm{d} t} \\& a=\frac{1}{\hbar}\left(\frac{d^2 E}{d k^2}\right) \frac{\mathrm{d} k}{\mathrm{~d} t}\end{aligned} .$

In quantum theory, the momentum of an electron is given by:

$p=\hbar k .$......(5)

Differentiating momentum with respect to t

$\frac{\mathrm{dp}}{\mathrm{d} t}=\hbar \frac{\mathrm{dk}}{\mathrm{d} t}$.......(6)

eq no (4)  i.e $a=\frac{1}{\hbar}\left(\frac{d^2 E}{d k^2}\right) \frac{\mathrm{dk}}{\mathrm{d} t}$ using eq no (6), reduces to

$a=\frac{1}{\hbar^2}\left(\frac{\mathrm{d}^2 \mathrm{E}}{\mathrm{d} k^2}\right) \frac{\mathrm{d} p}{\mathrm{~d} t}$.....(7)

using Newtons second law$\frac{\mathrm{d} p}{\mathrm{~d} t}=m^* a$ equation (6) reduces to

$a=\frac{1}{\hbar^2}\left(\frac{\mathrm{d}^2 \mathrm{E}}{\mathrm{d} k^2}\right) m^* a$

This mass of an electron is called the effective mass of an electron, denoted as m*.

The effective mass is thus determined by $d^2E/dK^2$.

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