A proton accelerated through a potential difference V has a de Broglie wavelength λ. On doubling the accelerating potential, the de Broglie wavelength of the proton:
Decreases
Understanding de Broglie Wavelength and Accelerating Potential
The de Broglie hypothesis states that all matter exhibits wave-like properties, and the wavelength associated with a particle is called the de Broglie wavelength (λ). This wavelength is inversely proportional to the momentum (p) of the particle.
The formula for de Broglie wavelength is given by:
\(\lambda = \frac{h}{p}\)
where \(h\) is Planck's constant.
Kinetic Energy from Accelerating Potential
When a charged particle, like a proton with charge \(q\), is accelerated through a potential difference \(V\), it gains kinetic energy (\(K\)). The kinetic energy gained is equal to the work done on the particle by the electric field:
\(K = qV\)
Connecting Kinetic Energy to Momentum
The kinetic energy of a particle with mass \(m\) and momentum \(p\) is also given by:
\(K = \frac{p^2}{2m}\)
We can rearrange this equation to express momentum in terms of kinetic energy:
\(p^2 = 2mK\)
\(p = \sqrt{2mK}\)
Deriving de Broglie Wavelength for an Accelerated Proton
Now, we can substitute the expression for kinetic energy (\(K = qV\)) into the momentum equation:
\(p = \sqrt{2m(qV)}\)
\(p = \sqrt{2mqV}\)
Substitute this expression for momentum into the de Broglie wavelength formula (\(\lambda = \frac{h}{p}\)):
\(\lambda = \frac{h}{\sqrt{2mqV}}\)
This formula shows how the de Broglie wavelength of a charged particle accelerated by a potential difference \(V\) depends on \(V\), its mass \(m\), and its charge \(q\). For a proton, \(m\) and \(q\) are constants.
Analyzing the Effect of Doubling Accelerating Potential
Initially, the proton is accelerated through potential \(V\), and its de Broglie wavelength is \(\lambda_1 = \frac{h}{\sqrt{2mqV}}\).
Now, the accelerating potential is doubled to \(V' = 2V\). Let the new de Broglie wavelength be \(\lambda_2\). Using the formula derived above:
\(\lambda_2 = \frac{h}{\sqrt{2mq(2V)}}\)
\(\lambda_2 = \frac{h}{\sqrt{2mqV \cdot 2}}\)
\(\lambda_2 = \frac{h}{\sqrt{2mqV} \cdot \sqrt{2}}\)
We can see that \(\frac{h}{\sqrt{2mqV}}\) is the original wavelength \(\lambda_1\). So,
\(\lambda_2 = \frac{\lambda_1}{\sqrt{2}}\)
Since \(\sqrt{2}\) is approximately 1.414, which is greater than 1, the new wavelength \(\lambda_2\) is smaller than the original wavelength \(\lambda_1\).
Specifically, the new wavelength is \(\frac{1}{\sqrt{2}}\) times the original wavelength.
Therefore, on doubling the accelerating potential, the de Broglie wavelength of the proton decreases.
| Parameter | Initial State | Final State (Potential Doubled) | Change |
|---|---|---|---|
| Accelerating Potential (\(V\)) | \(V\) | \(2V\) | Doubled |
| Kinetic Energy (\(K=qV\)) | \(qV\) | \(q(2V) = 2qV\) | Doubled |
| Momentum (\(p=\sqrt{2mK}\)) | \(\sqrt{2mqV}\) | \(\sqrt{2m(2qV)} = \sqrt{2}\sqrt{2mqV}\) | Increases by factor of \(\sqrt{2}\) |
| de Broglie Wavelength (\(\lambda=\frac{h}{p}\)) | \(\frac{h}{\sqrt{2mqV}}\) | \(\frac{h}{\sqrt{2}\sqrt{2mqV}} = \frac{1}{\sqrt{2}} \times \frac{h}{\sqrt{2mqV}}\) | Decreases by factor of \(\frac{1}{\sqrt{2}}\) |
Conclusion on de Broglie Wavelength Change
When the accelerating potential difference for a proton is doubled, the kinetic energy it gains also doubles. Since momentum is proportional to the square root of kinetic energy, the momentum increases by a factor of \(\sqrt{2}\). As de Broglie wavelength is inversely proportional to momentum, the wavelength decreases by a factor of \(\frac{1}{\sqrt{2}}\).
Thus, the de Broglie wavelength of the proton decreases.
Revision Table: de Broglie Wavelength Concepts
| Concept | Formula | Relationship to Momentum | Relationship to Potential \(V\) (for charged particle) |
|---|---|---|---|
| de Broglie Wavelength (\(\lambda\)) | \(\lambda = \frac{h}{p}\) | Inversely proportional (\(\lambda \propto \frac{1}{p}\)) | Inversely proportional to square root (\(\lambda \propto \frac{1}{\sqrt{V}}\)) |
| Momentum (\(p\)) | \(p = mv\) or \(p = \sqrt{2mK}\) | Inversely proportional to \(\lambda\) | Proportional to square root (\(p \propto \sqrt{V}\)) |
| Kinetic Energy (\(K\)) | \(K = \frac{1}{2}mv^2\) or \(K = qV\) | Proportional to \(p^2\) | Proportional (\(K \propto V\)) |
Additional Information: Applications of de Broglie Wavelength
The wave-like nature of particles, described by the de Broglie wavelength, is a fundamental concept in quantum mechanics. This principle has important practical applications:
- Electron Microscopy: Electron microscopes utilize the wave properties of electrons. Because the de Broglie wavelength of electrons accelerated through high potentials can be much smaller than the wavelength of visible light, electron microscopes can achieve much higher resolution than optical microscopes, allowing us to see much finer details of materials.
- Neutron Diffraction: Neutrons also exhibit wave properties. Neutron diffraction is a technique used to study the atomic and magnetic structure of materials, similar to how X-ray diffraction is used, but providing different information due to the neutrons' properties (e.g., magnetic moment, non-charge).
- Quantum Computing: Understanding and manipulating the wave nature of particles is essential for developing quantum computers and other quantum technologies.
The de Broglie wavelength is a key aspect of wave-particle duality, demonstrating that particles can behave like waves and waves can behave like particles.
Important Questions from Dual Nature of Radiation and Matter
A point source causing photoelectric emission from a metallic plate is moved away from the plate. The variation of photoelectric current with distance from the source is correctly represented by the graph:
The wavelength of light from the spectral emission line of sodium is 662 nm. The kinetic energy at which an electron would have the same de Broglie wavelength would be: (h=6.62×10−34J⋅s,me=9×10−31kg)
Energy of a photon corresponding to a wavelength of 600 nm is 2.08 eV. The energy of a photon of wavelength 400 nm will be:
A particle moves three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. The mass of the particle is:
According to Einstein’s photoelectric equation, the plot of the Kinetic Energy of the emitted photoelectrons from a metal versus the frequency of the incident radiation gives a straight line whose slope:
