Chapter 10: Q61E (page 470)

Question: - Verify using equation (10-12) that the effective mass of a free particle is m.

Expert verified

Answer: -

The effective mass of a free particle is $m = h^{2} {(\frac{d^{2} E}{d k^{2}})}^{- 1}$ .

Step by step solution

The relation between the momentum p of a particle and a wavenumber is, where is a reduced plank constant.

The kinetic energy E of the particle is related to the momentum as:

$E = \frac{p^{2}}{2 m}$

Substitute the value of P as h_k. Thus,

$E = \frac{{(h k)}^{2}}{2 m} E = \frac{h^{2} k^{2}}{2 m}$

Differentiate the above equation with respect to K.

$\frac{d E}{d k} = \frac{d}{d k} (\frac{h^{2} k^{2}}{2 m}) = \frac{2 h^{2} k}{2 m} = \frac{h^{2} k}{m}$

Differentiate the above equation again with respect to K.

$\frac{d^{2} E}{d k^{2}} = \frac{d}{d k} (\frac{h^{2} k}{m}) = \frac{h^{2}}{m}$

The effective mass is calculated by rearranging the above equation for :

$m = h^{2} {(\frac{d^{2} E}{d k^{2}})}^{- 1}$

Hence, The effective mass of a free particle is $m = h^{2} {(\frac{d^{2} E}{d k^{2}})}^{- 1}$ .

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