Determine the de Broglie wavelength for a 54-eV electron used by Davisson and Germer.

Strategy We shall use the de Broglie wavelength Equation (5.2) to determine the wavelength λ. We need to find the momentum of a 54-eV electron, but because the energy is so low, we can do a nonrelativistic calculation. We shall do a general calculation for the wavelength of any electron accelerated by a voltage of $V_{0}$ .

$\lambda=\frac{h}{p}$ (5.2)

Question

Determine the de Broglie wavelength for a 54-eV electron used by Davisson and Germer.

Strategy We shall use the de Broglie wavelength Equation (5.2) to determine the wavelength λ. We need to find the momentum of a 54-eV electron, but because the energy is so low, we can do a nonrelativistic calculation. We shall do a general calculation for the wavelength of any electron accelerated by a voltage of [latex]V_{0}[/latex].

[latex]\lambda=\frac{h}{p}[/latex] (5.2)

Accepted Answer

We write the kinetic energy K.E. in terms of the final momentum of the electron and the voltage [latex]V_{0}[/latex] across which the electron is accelerated.

[latex]\frac{p^{2}}{2 m}= K . E .=e V_{0}[/latex] (5.6)

We find the momentum from this equation to be [latex]p=\sqrt{(2 m)\left(e V_{0} ight)}[/latex]. The de Broglie wavelength from Equation (5.2) is now

[latex]\begin{aligned}\lambda=\frac{h}{p}=\frac{h c}{p c} &=\frac{h c}{\sqrt{\left(2 m c^{2} ight)\left(e V_{0} ight)}} \&=\frac{1240 eV \cdot nm }{\sqrt{(2)\left(0.511 imes 10^{6} eV ight)\left(e V_{0} ight)}}\end{aligned}[/latex]

[latex]\lambda=\frac{1.226 nm \cdot V ^{1 / 2}}{\sqrt{V_{0}}}[/latex] (5.7)

where the constants h, c, and m have been evaluated and [latex]V_{0}[/latex] is the voltage. For [latex]V_{0}=54[/latex] V, the wavelength is

[latex]\lambda=\frac{1.226 nm \cdot V ^{1 / 2}}{\sqrt{54 V }}=0.167 nm[/latex]

Question 5.3: Determine the de Broglie wavelength for a 54-eV electron use...

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