Question

What is the de Broglie wavelength of an electron traveling at $\text{2}\text{.0 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{8}}}\text{ m}{{\text{s}}^{-1}}?$

Answer 1

Matter waves, which are an example of wave–particle duality, are an important element of quantum mechanics theory. All stuff behaves in a wavelike manner. A beam of electrons, for example, can be diffracted in the same way as a beam of light or a water wave can. However, in most situations, the wavelength is too short to have a practical effect on daily tasks. In 1924, French scientist Louis de Broglie introduced the idea that matter acts like a wave. The de Broglie theory is another name for it. De Broglie waves are the name given to matter waves.

Complete answer:
The de Broglie wavelength, $\lambda$ , is linked with a substantial particle (as opposed to a massless particle) and is connected to its momentum, p, via the Planck constant, h: $\lambda =\dfrac{h}{p}=\dfrac{h}{mv}$
George Paget Thomson's thin metal diffraction experiment and the Davisson–Germer experiment, both employing electrons, were the first to reveal wave-like behaviour of matter, and it has since been verified for other elementary particles, neutral atoms, and even molecules. Its value is the same as the Compton wavelength when c = v.
The De Broglie wavelength is a wavelength exhibited in all objects in quantum mechanics that defines the probability density of locating the item at a particular position in the configuration space, according to wave-particle duality. The momentum of a particle is inversely related to its de Broglie wavelength.
$\lambda=3.64 \cdot 10^{-12} \mathrm{~m}$
Explanation:
de Broglie wave equation is given as $\rightarrow \lambda=\dfrac{h}{p}$
where
- $\lambda$ denotes the wavelength in $\mathrm{m}$ .
- $p(\operatorname{mass}(m) \cdot \operatorname{velocity}(v))$ denotes momentum
(electron mass $=9.109 \cdot 10^{-31} \mathrm{~kg}$ )
- $h$ denotes Planck's constant $=6.626 \cdot 10^{-34} \mathrm{~J}($ joule $) \cdot \mathrm{s}($ second)
$\left(1 \text { Joule }=1 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}^{2}\right)$
On solving
$\lambda =\dfrac{6.626\cdot {{10}^{-34}}~\text{J}\cdot \text{s}}{mv}$
$\lambda =\dfrac{6.626\cdot {{10}^{-34}}~\text{J}\cdot \text{s}}{\left( 9.109\cdot {{10}^{-31}}~\text{kg} \right)\left( 2.0\cdot {{10}^{8}}~\text{m}{{\text{s}}^{-1}} \right)}$
$\lambda =\dfrac{6.626\cdot {{10}^{-34}}~\text{kg}\cdot {{\text{m}}^{2}}{{\text{s}}^{-1}}}{18.2\cdot {{10}^{-23}}~\text{kg}\cdot \text{m}{{\text{s}}^{-1}}}$ Here, everything is cancelled except $\mathrm{m}$
$\Rightarrow \lambda =3.64\cdot {{10}^{-12}}~\text{m}$

Note:
The de Broglie waves exist as a closed-loop in the case of electrons travelling in circles around the nuclei in atoms, thus they can only exist as standing waves and fit evenly around the loop. As a result of this need, atoms' electrons orbit the nucleus in specific configurations, or states, known as stationary orbits.