Calculate the change in kinetic energy of an electron when the velocity changes by a small amount. Consider an electron traveling at a velocity of 10^7 cm/s. Assume that the velocity increases by a value of 1 cm/s. The increase in kinetic energy is given by ΔE = 1/2 mv2^2 - 1/2 mv1^2 = 1/2 m(v2^2 -

Question

Calculate the change in kinetic energy of an electron when the velocity changes by a small amount.

Consider an electron traveling at a velocity of [latex]10^{7} \mathrm{~cm} / \mathrm{s}[/latex]. Assume that the velocity increases by a value of [latex]1 \mathrm{~cm} / \mathrm{s}[/latex]. The increase in kinetic energy is given by

[latex] \Delta E=\frac{1}{2} m v_{2}^{2}-\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m\left(v_{2}^{2}-v_{1}^{2}\right) [/latex]

Let [latex]v_{2}=v_{1}+\Delta v[/latex]. Then

[latex] v_{2}^{2}=\left(v_{1}+\Delta v\right)^{2}=v_{1}^{2}+2 v_{1} \Delta v+(\Delta v)^{2} [/latex]

But [latex]\Delta v \ll v_{1}[/latex], so we have that

[latex] \Delta E \approx \frac{1}{2} m\left(2 v_{1} \Delta v\right)=m v_{1} \Delta v [/latex]

Accepted Answer

Substituting the number into this equation, we obtain

[latex] \Delta E=\left(9.11 imes 10^{-31} ight)\left(10^{5} ight)(0.01)=9.11 imes 10^{-28} \mathrm{~J} [/latex]

which may be converted to units of electron volts as

[latex] \Delta E=\frac{9.11 imes 10^{-28}}{1.6 imes 10^{-19}}=5.7 imes 10^{-9} \mathrm{eV} [/latex]

Comment

A change in velocity of [latex]1 \mathrm{~cm} / \mathrm{s}[/latex] compared with [latex]10^{7} \mathrm{~cm} / \mathrm{s}[/latex] results in a change in energy of [latex]5.7 imes 10^{-9} \mathrm{eV}[/latex], which is orders of magnitude larger than the change in energy of [latex]10^{-19} \mathrm{eV}[/latex] between energy states in the allowed energy band. This example serves to demonstrate that a difference in adjacent energy states of [latex]10^{-19} \mathrm{eV}[/latex] is indeed very small, so that the discrete energies within an allowed band may be treated as a quasi-continuous distribution.

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