Example 6.6: Calculating the Wavelength of a Particle

If an electron travels at a velocity of \(1.000\times 10^{7 }\)m/s and has a mass of \(9.109\times 10^{-28 }\)g, what is its wavelength?

Solution

\(v_{\mathrm{elec}}\) \(= 1.000\times 10^{7}\ \frac{\mathrm{m}}{\mathrm{s}}\)


\(m_{\mathrm{elec}}\) \(= 9.109\times 10^{-28}\ \mathrm{g}\)


We can use de Broglie’s equation to solve this problem

\(λ = \dfrac{h_{\mathrm{Planck}}}{m \cdot v}\)     

\(h_{\mathrm{Planck}}\) \(= 6.626\times 10^{-34}\ \mathrm{J}\ \mathrm{s}\)


\(λ_{\mathrm{elec}}\) \(= \dfrac{h_{\mathrm{Planck}}}{m_{\mathrm{elec}} \cdot v_{\mathrm{elec}}}\)

\(\ \ \ =\dfrac{6.626\times 10^{-34}\ \mathrm{J}\ \mathrm{s}}{9.109\times 10^{-28}\ \mathrm{g} \cdot 1.000\times 10^{7}\ \frac{\mathrm{m}}{\mathrm{s}}}\)

\(\ \ \ =\dfrac{6.626\times 10^{-34}\ \mathrm{J}\ \mathrm{s}}{9.1090\times 10^{-21}\ \frac{\mathrm{g}\ \mathrm{m}}{\mathrm{s}}}\)

\(\ \ \ =7.274\times 10^{-11}\ \mathrm{m}\)


\(λ_{\mathrm{elec}}\) \(= 7.274\times 10^{-11}\ \mathrm{m}\)

\(\ \ \ =0.7274\ \mathrm{Å}\)


This is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom.