If what speed is an electron&qpos;s de Broglie wavelength

(a) $1.0 \mathrm{pm}$ , (b) $1.0 \mathrm{~nm}$ , (c) $1.0 \mu \mathrm{m}$ , and (d) $1.0 \mathrm{~mm}$ ?

(a) $1.0 \mathrm{pm}$ , (b) $1.0 \mathrm{~nm}$ , (c) $1.0 \mu \mathrm{m}$ , and (d) $1.0 \mathrm{~mm}$ ?

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38.16. Solve: (a) The de Broglie wavelength is\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) v}=1.0 \times 10^{-12} \mathrm{~m} \Rightarrow v=7.3 \times 10^{8} \mathrm{~m} / \mathrm{s}This speed is larger than c, indicating a breakdown of de Broglie's equation. This is an acceptable answer if you haven't studied relativity. However, a better approach would be to use the relativistic form for the momentum, p=\gamma m v. Hence,\begin{aligned}\lambda & =\frac{h}{\gamma m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s} \sqrt{1-v^{2} / c^{2}}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) v}=1.0 \times 10^{-12} \mathrm{~m} \\\Rightarrow \sqrt{1-v^{2} / c^{2}} & =\left(\frac{v}{c}\right) \frac{\left(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\left(1.0 \times 10^{-12} \mathrm{~m}\right)\left(9.11 \times 10^{-31} \mathrm{~kg}\right)}{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}=0.4122\left(\frac{v}{c}\right) \\& \Rightarrow 1-\frac{v^{2}}{c^{2}}=0.170 \frac{v^{2}}{c^{2}} \Rightarrow v=0.925 c=2.8 \times 10^{8} \mathrm{~m} / \mathrm{s}\end{aligned}(b) For \lambda=1.0 \times 10^{-9} \mathrm{~m},\lambda=1.0 \times 10^{-9} \mathrm{~m}=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}}{\left(9.11 \times 10^{-31} \mathrm{~kg}\right) v} \Rightarrow v=7.3 \times 10^{5} \mathrm{~m} / \mathrm{s}(c) Likewise, for \lambda=1.0 \times 10^{-6} \mathrm{~m}, v=7.3 \times 10^{2} \mathrm{~m} / \mathrm{s}.(d) For \lambda=1.0 \times 10^{-3} \mathrm{~m}, v=0.73 \mathrm{~m} / \mathrm{s}. ...