Engineering Physics

LESSON 17. Effective Mass and Mathematical Derivation

\[v={{2\pi } \over h}{{dE} \over {dk}}\]……………….(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 \[{{{d^2}E} \over {d{k^2}}}\]

For free electron m* = m  because

\[E={{{\hbar ^2}{k^2}} \over {2m}}\]  and

\[{{{d^2}E} \over {d{k^2}}} = {{{\hbar ^2}} \over m}\]

This result is expected, but for an electron moving in a periodic potential, E dose not vary with κ in the above manner, and so m* ≠ m. Thus we conclude that all the results of the free electron theory are correct provided m  in each case is replaced by a suitable m* .

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