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Show that the de Broglie wavelength of an electron accelerated from rest through a small potential difference \(V\) is given by \(\lambda=1.226 / \sqrt{V}\), where \(\lambda\) is in nanometers and \(V\) is in volts.

Short Answer

Expert verified
The de Broglie wavelength of an electron accelerated from rest through a potential difference \(V\) can be calculated using the principles of the de Broglie hypothesis and the conservation of energy principle. By applying these principles and substituting in the real-world values, we have shown that the de Broglie wavelength is indeed given by \(\lambda = \frac{1.226}{\sqrt{V}}\) where \(\lambda\) is in nanometers and \(V\) is in volts.

Step by step solution

01

Understand De Broglie Wavelength

De Broglie wavelength of a particle is given by \(\lambda = \frac{h}{p}\) where, \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant and \(p\) is the momentum of the particle. In our context, the particle is an electron accelerated through a potential difference.
02

Define Momentum for an Electron

The momentum of a particle is given by \(p = \sqrt{2m_eE}\) where, \(m_e\) is the mass of an electron and \(E\) is the kinetic energy.
03

Calculate Kinetic Energy

The kinetic energy of an electron accelerated through a potential difference \(V\) is given by \(E = eV\) where, \(e\) is the charge of the electron.
04

Substitute E from step 3 to step 2

Substituting the value of \(E\) from step 3 into the equation for \(p\) in step 2, we get \(p = \sqrt{2m_eeV}\).
05

Substitute p from step 4 to step 1

Now, substituting the value of \(p\) from step 4 into the de Broglie wavelength formula: \(\lambda = \frac{h}{\sqrt{2m_eeV}}\).
06

Final Simplification

After converting all above mentioned quantities to their respective SI units and inputting their respective values (Planck's constant \(h = 6.62607015 x 10^{-34} m^2kg/s\), electron mass \(m_e = 9.10938356 × 10^−31 kg\), and electron charge \(e = 1.602 x 10^{-19} C\)), the above equation simplifies to \(\lambda = \frac{1.226}{\sqrt{V}}\) when \(\lambda\) is in nanometers and V is in volts, which is the required proof.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantum Mechanics and the De Broglie Wavelength
Quantum mechanics, a fundamental theory in physics, provides a comprehensive framework for understanding the behavior of particles at the nanoscopic scale. This field fundamentally differs from classical physics, as it incorporates principles such as wave-particle duality and quantization of energy.

One of the key ideas in quantum mechanics is the concept of the de Broglie wavelength. Postulated by Louis de Broglie, it is the wave property associated with particles that have mass. It articulates how every moving particle or object has an associated wavelength, which can be calculated using the formula \(\lambda = \frac{h}{p}\), where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the particle's momentum.

Understanding this concept is crucial for interpretative scenarios such as electron diffraction and the operation of electron microscopes. These applications stem from the recognition of electrons behaving not just as particles, but also as waves that can interfere and create patterns typical of wave phenomena.
Electron Momentum and Its Significance
Momentum in quantum mechanics is a delicate concept; it represents the product of the mass and velocity of a particle but encapsulated within the wave-like nature of the particle in context. For an electron, or any particle with mass, the momentum \(p\) plays a pivotal role in defining its de Broglie wavelength.

Considering an electron, its momentum is determined by the equation \(p = \sqrt{2m_eE}\), with \(m_e\) being the mass of the electron and \(E\) its kinetic energy. When an electron is accelerated — such as through a potential difference \(V\) — its kinetic energy changes, subsequently affecting its momentum.

The relationship of kinetic energy to the potential difference, \(E = eV\), where \(e\) is the charge of the electron, illustrates how the electron speed and thus its momentum can be controlled through an applied voltage. This forms the basis for numerous electron-based technologies, such as particle accelerators and imaging instruments.
Potential Difference and Its Effect on Electron Behavior
The concept of potential difference is a cornerstone in electromagnetism and, by extension, quantum mechanics when it deals with charged particles like electrons. A potential difference \(V\), typically measured in volts, is essentially the work done per unit charge to move a charge between two points in an electric field.

Applying a potential difference causes an electron, initially at rest, to accelerate and thereby gain kinetic energy quantified by \(E = eV\), where \(e\) is the fundamental electron charge. This transformation connects classical electromagnetic concepts with quantum mechanical behavior, as the resulting electron movement can be described using quantum principles.

The effect of a potential difference on electron behavior is profound. By accelerating electrons, we can effectively control their de Broglie wavelength and thereby manipulate their wave-like interactions. This precise adjustment is what enables applications like the calibration of the electron beam in a scanning electron microscope, where the de Broglie wavelength is directly linked to the resolution of the image.

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Most popular questions from this chapter

To "observe" small objects, one measures the diffraction of particles whose de Broglie wavelength is approximately equal to the object's size. Find the kinetic energy (in electron volts) required for electrons to resolve (a) a large organic molecule of size \(10 \mathrm{~nm}\), (b) atomic features of size \(0.10 \mathrm{~nm}\), and \((\mathrm{c})\) a nucleus of size \(10 \mathrm{fm}\). Repeat these calculations using alpha particles in place of electrons.

An excited nucleus with a lifetime of \(0.100 \mathrm{~ns}\) emits a \(\gamma\) ray of energy \(2.00 \mathrm{MeV}\). Can the energy width (uncertainty in energy, \(\Delta E\) ) of this \(2.00-\mathrm{MeV} \gamma\) emission line be directly measured if the best gamma detectors can measure energies to \(\pm 5 \mathrm{eV}\) ?

A monoenergetic beam of electrons is incident on a single slit of width \(0.50 \mathrm{~nm} .\) A diffraction pattern is formed on a screen \(20 \mathrm{~cm}\) from the slit. If the distance between successive minima of the diffraction pattern is \(2.1 \mathrm{~cm}\), what is the energy of the incident electrons?

Find the de Broglie wavelength of a ball of mass \(0.20 \mathrm{~kg}\) just before it strikes the Earth after being dropped from a building \(50 \mathrm{~m}\) tall.

Suppose Fuzzy, a quantum-mechanical duck, lives in a world in which \(h=2 \pi \mathrm{J} \cdot \mathrm{s}\). Fuzzy has a mass of \(2.0 \mathrm{~kg}\) and is initially known to be within a region \(1.0 \mathrm{~m}\) wide. (a) What is the minimum uncertainty in his speed? (b) Assuming this uncertainty in speed to prevail for \(5.0 \mathrm{~s}\), determine the uncertainty in position after this time.

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