Exercise of de Broglie wave

In quantum mechanics, we believe that every object has both particle and wave properties. If so, there should be rules for linking the properties of the particles and those of the waves.

De Broglie suggested the following equation, thinking that particles with masses such as protons and electrons can also be regarded as waves.


\(h\) is Planck’s constant, \(p\) is the particle momentum. Then the particle has the same properties as wave of wavelength \(λ\). This wavelength \(λ\) is called the de Broglie wavelength.

Case1 A Large Object

Exercise: Calculate the De Broglie’s wavelength of a person with a mass of 50.0 kg who is moving at a speed of 10.0 km/h.

First, fix the unit of speed 10.0 km/h to m/s.


Let Plank’s constant be \(h=6.63×10^{-34}[J·s]\).


From this, in the macro world we are aware of, the de Broglie wavelength comes up considerably shorter.

Case2 An Electron

Suppose that the observed de Broglie’s wavelength of an electron is \(5.0×10^{-11}[m] \). Calculate the acceleration voltage of the electron.

Assuming that the charge of an electron is \(e\), the energy \(E\) that the electron obtain from the applied voltage \(V\) is expressed by the following formula.


Here, the mass of the electron is \(m_e\), and the speed and momentum of the electron are \(v\) and \(p\) respectively, the following expression holds.

$$|e|V=\frac{m_e v^2}{2}=\frac{p^2}{2m_e}$$

$$p=\sqrt{2m_e |e|V}$$

Substitute this equation into the de Broglie’s wavelength equation.

$$5.0×10^{-11}=\frac{6.63×10^{-34}}{\sqrt{2m_e |e|V}}$$

Therefore, the voltage \(V\) is calculated.

\begin{eqnarray} V&=&\left( \frac{6.63×10^{-34}}{5.0×10^{-11}}\right)^2 \frac{1}{2m_e |e|}\\&=&\left( \frac{6.63×10^{-34}}{5.0×10^{-11}}\right)^2 \frac{1}{2・9.11×10^{-31}・ 1.60×10^{-19}}\\&≒&603[V]\end{eqnarray}

In conclusion, if the electron is applied the acceleration voltage \(603V\), the de Broglie wavelength of the electron becomes \(5.0×10^{- 11}[m]\).


・De Broglie’s idea makes it easier to express that an object has both particle and wave properties.

・ The larger the momentum of the object, the shorter the De Broglie wavelength becomes and vice versa.

Leave a Reply