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The concept of the de Broglie wavelength is central to understanding wave-particle duality. Louis de Broglie proposed that every moving particle or object has associated with it a wave nature. This is famously represented by the de Broglie equation: \[ \lambda = \frac{h}{p} \]Where:

• \( \lambda \) is the de Broglie wavelength, the wavelength associated with a particle.
• \( h \) is Planck's constant, approximately \( 6.63 \times 10^{-34} \mathrm{~Js} \).
• \( p \) is the momentum of the particle, given by \( p = mv \), where \( m \) is mass and \( v \) is velocity.

In the context of the exercise, to "see" or resolve atomic structures that are about \(0.1 \mathrm{~nm}\) in size, an electron must have a wavelength at least that small. This necessitates calculating the momentum, and subsequently, the velocity of the electron required to achieve such a small de Broglie wavelength.

Resolution at the atomic scale requires radiation of wavelengths similar to the size of the atoms. Imagine trying to see atoms—tiny fractions of matter about \(0.1 \mathrm{~nm}\) in size. To effectively visualize such small structures, the resolving tool (in this case, an electron) must similarly have a very tiny associated wavelength.Using the de Broglie wavelength formula, if we substitute the target atomic size \( \lambda = 0.1 \times 10^{-9} \mathrm{~m} \), we gather insight into what is demanded of the kind of particle that can "see" this scale.This scale is crucial for technologies like electron microscopy, where electrons are employed due to their ability to achieve such fine resolutions, unlike visible light, which is inadequate for resolving such minute details because of its comparatively larger wavelength.

Once we establish that an electron must have a de Broglie wavelength of approximately \(0.1 \mathrm{~nm}\) for observing atomic structures, we need to calculate how fast it must travel to achieve this.From the de Broglie relation, we have:\[ p = \frac{h}{\lambda} \]To find the velocity \( v \), remember that the momentum \( p \) of the electron is also expressed as \( p = m_e v \). Rearranging gives:\[ v = \frac{p}{m_e} \]Substitute the known values:

• \( h = 6.63 \times 10^{-34} \mathrm{~Js} \)
• \( \lambda = 0.1 \times 10^{-9} \mathrm{~m} \)
• \( m_e = 9.11 \times 10^{-31} \mathrm{~kg} \)

Calculate \( p \) using the de Broglie relation:\[ p = \frac{6.63 \times 10^{-34}}{0.1 \times 10^{-9}} \]Subsequently, find the electron's velocity:\[ v = \frac{p}{9.11 \times 10^{-31}} \]Ultimately, this calculation provides the speed necessary for the electron to have a wavelength capable of "seeing" an atom, emphasizing how quantum mechanics can offer profound insights into the natural world.