Question

At what speed is an electron's de Broglie wavelength (a) $1.0\mathrm{pm}$, (b) $1.0\mathrm{nm}$, (c) $1.0\mu \mathrm{m}$, and (d)$1.0\mathrm{mm}$ ?

Short answer

(a) $v=7.27\times {10}^{8}m/s$

(b) $v=2.77\times {10}^{8}m/s$

(c) $v=7.27\times {10}^{2}m/s$

(d)$v=0.727m/s$

Step-by-step solution

Part(a) Step 1: Given Information 

The electron's de Broglie wave length is$1.0pm$

Part(a) Step 2: Calculation

We can start the solution with expression for de-Broglie wavelength

$$\begin{gathered} \lambda =\frac{h}{mv} \\ v=\frac{h}{m\lambda } \\ v=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(1\times {10}^{-12}\right)} \\ v=7.27\times {10}^{8}m/s \end{gathered}$$

This speed is larger than speed of light, so we must calculate relativistic speed:

$$\begin{gathered} \lambda =\frac{h}{p} \\ p=\gamma mv \\ \frac{h}{\lambda }=\gamma mv \\ \gamma =\frac{1}{\frac{h}{1-\frac{v^2}{c^2}}} \\ \frac{h}{\lambda }\left(1-\frac{v^2}{c^2}\right)=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}m{v}^2 \\ \frac{h^2}{{\lambda }^2}=v^2\left(m^2+\frac{h^2}{{\lambda }^2{c}^2}\right) \\ v=\frac{h}{\lambda ·\sqrt{\left(m^2+\frac{h^2}{{\lambda }^2{c}^2}\right)}} \\ v=2.77\times {10}^{8}m/s \end{gathered}$$

Part(b) Step 1: Given Information 

The electron's de Broglie wave length is$1.0nm$

Part(b) Step 2: Calculation 

For $\lambda =1\times {10}^{-9}$we have:

$$\begin{gathered} \lambda =\frac{h}{mv} \\ v=\frac{h}{m\lambda } \\ v=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(1\times {10}^{-9}\right)} \\ v=7.27\times {10}^{5}m/s \end{gathered}$$

Part(c) Step 1: Given Information 

The electron's de Broglie wave length is$1.0\mu m$

Part(c) Step 2: Calculation

For $\lambda =1\times {10}^{-6}$we have:

$$\begin{gathered} v=\frac{h}{m\lambda } \\ v=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(1\times {10}^{-6}\right)} \\ v=7.27\times {10}^{2}m/s \end{gathered}$$

Part(d) Step 1: Given Information 

The electron's de Broglie wave length is$1.0mm$

Part(d) Step 2: Calculation

For $\lambda =1\times {10}^{-3}$we have:

$$\begin{gathered} v=\frac{h}{m\lambda } \\ v=\frac{6.63\times {10}^{-34}}{\left(9.11\times {10}^{-31}\right)\left(1\times {10}^{-3}\right)} \\ v=0.727m/s \end{gathered}$$