LESSON 17. Effective Mass and Mathematical Derivation

Effective Mass

• We know that an electron has a well defined mass, and it obey Newtonian Mechanics when accelerated by an electric field  E.

• Now in the situation of an electron moving inside a crystal i.e. in initial state k, where it is not free to move, the question arises as to what the mass of an electron inside crystal is under accelerated motion and what is the effect of an electric field?

• It is found that the mass of and electron in a crystal appears, in general, different from the free electron mass, and is usually referred to as the effective mass.

• The velocity of the electron in a one-dimensional lattice is described by its group velocity given by

$v={{2\pi } \over h}{{dE} \over {dk}}$……………….(1)

and acceleration is given by

$a={{dv} \over {dt}}=\left( {{{2\pi } \over h}} \right){d \over {dt}}\left( {{{dE} \over {dk}}} \right)$

$a=\left( {{{2\pi } \over h}} \right)\left( {{{{d^2}E} \over {d{k^2}}}} \right){{dk} \over {dt}}$………………..(2)

Now E and k relation,

We can obtain ${{{d^2}E} \over {d{k^2}}}$  . Now under the influence of an applied electric field E we have to fiend the value of ${{dk} \over {dt}}$.

Consider that an electron is subjected to an external electric field E for a time  dt

If the velocity of the electron is ν   and the distance travelled in time dt is ν dt

$dE=eEvdt$……………….(4)

Substituting the value of equation (1) in equation (4)

$dE=eE{{2\pi } \over h}{{dE} \over {dk}}dt$………..(5)

${{dk} \over {dt}}$ = ${{2\pi eE} \over h}$.................(6)

Substitute of equation (6) in equation (2)

$a=\left( {{{2\pi } \over h}} \right)\left( {{{{d^2}E} \over {d{k^2}}}} \right){{2\pi eE} \over h}$

Hence

$a=\left( {{{4{\pi ^2}} \over {{h^2}}}} \right)eE\left( {{{{d^2}E} \over {d{k^2}}}} \right)$………..(7)

Comparing equation (7) with that for a free, classical particle

F = ma, $F=m{{dv} \over {dt}}$   and  F = eE

Hence

ma = $m{{dv} \over {dt}}$ = eE

$a={{eE} \over m}$ ……….(8)

From eqn(7) and (8)

${{eE} \over m}$ =  $\left( {{{4{\pi ^2}} \over {{h^2}}}} \right)eE\left( {{{{d^2}E} \over {d{k^2}}}} \right)$

Therefore the effective mass is given by

Hence,

It indicates the effective mass of an electron moving through the crystal lattice is determine by ${{{d^2}E} \over {d{k^2}}}$

$E={{{\hbar ^2}{k^2}} \over {2m}}$  and
${{{d^2}E} \over {d{k^2}}} = {{{\hbar ^2}} \over m}$
Therefore,$E={{{\hbar ^2}{k^2}} \over {2{m^*}}}$ is known as effective mass approximation.