Question regarding the de Broglie wavelength of a particle.

• Simen
Set the total energy E equal to the sum of the rest energy (mc^2) and the kinetic energy (eV).2. Set E equal to the relativistic energy (sqrt(p^2c^2+m^2c^4)).3. Solve for p in terms of e and V.4. Use the de Broglie relation (p=h/lambda) to solve for lambda in terms of m, e, and V.Hope that helps!In summary, to find the de Broglie wavelength of a particle with charge e and mass m_0 accelerated by a potential V to a relativistic speed, we must first set the total energy E equal to the sum of the rest energy
Simen

Homework Statement

A particle with charge e and mass m_0 is accelerated by a potential V to a relativistic speed.

Show that the de Broglie wavelength is:

λ = h / √(2 m_0 eV ) √(1+( eV / 2 m_0 c^2 ))

Homework Equations

E = qV
λ = h/p
E = √(p^2 * c^2 + m^2 * c^4 )
p = (m_0 * v) / (1 - (v^2/c^2)

The Attempt at a Solution

We do not have an expression for v, and so we use the formula:
E = √(p^2 * c^2 + m^2 * c^4 )
which gives us that:
E^2/c^2 = p^2 +m^2 *e^2
Which gives:
p = √(E^2/c^2 + m^2 * c^2)
Putting p into the expression for the de Broglie wavelenght gives us:

λ = h/ √(E^2 / c^2 + m^2*c^2)

We know that the potential energy of the particle in the Field before accelleration starts (and therefore the kinetic energy) is simply eV, and put that into the expression as well.

λ = h/ √(eV^2 / c^2 + m^2*c^2)

It has been a very long time since I last did any real maths, super basic algebra was sufficient for most of what I've done for the last two years. I am guessing the path to making the expressions the same lies in derivation, though why that should be an operation I have no idea.

I am well and truly lost on this, so if anyone have any insight whatsoever, I will be very grateful.
(Oh, and this is one of several attempts down different roads. I think this is the one that is closest to a proper solution, and also the one that is the least confusing for those of you who actually know this stuff...)

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The kinetic energy of the particle is eV. The total energy E is the sum of the kinetic and rest energies. If you fix that, you'll probably see how to get it to work out.

1 person
Thank you so much for your reply, I'm still not entirely sure about how to og about this. I think the formula used for the energy in my post above is the formula for the kinetic energy, and the formula for the total energy would be:
Ek + mc2 = √(p^2 * c^2 + m^2*c^4)

which turns into:

Ek = √(p^2 * c^2 + m^2*c^4) - mc2
Ek 2 = p2 * c2 + 2m2*c4
Ek 2 / c2 = p2 +2m2 * c2
p2 = Ek 2 / c2 - 2m2 * c2

p = Ek / c - √(2m2 * c2)
or
p = √(eV2 / c2 -2m2 c2)

λ = h / √(eV2 / c2 -2m2 c2)

Is there a difference in the two masses? Should I put in a Lorentz factor somewhere to account for rest mass vs. mass at relativistic speed?

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I think, here we have to give some general concepts, before helping to solve the problem. It is very important to get the right idea about the quantities involved in relativistic particle physics, particularly not to confuse the students with unnecessary old ideas about "relativistic mass" etc. (see Simen's previous posting concerning his confusion/uncertainty about possible Lorentz factors!).

The de Broglie wavelength for a particle with momentum $p$ is given via the wave number of the corresponding wave mode in quantum field theory, i.e.,
$$k=\frac{p}{\hbar}=\frac{2 \pi}{\lambda} \; \Rightarrow \; \lambda=\frac{2 \pi \hbar}{p}=\frac{h}{p}.$$

For the given question, the total energy of the particle is given by
$$E=m_0 c^2 + |e V|=\sqrt{m_0^2 c^4 + p^2 c^2}, \qquad (1)$$
because the kinetic energy gained when a charged particle is accelerated starting at rest running through a electrostatic potential difference $V$ is given by $|e V|$, and in relativistic physics we always include the "rest energy" $m_0c^2$ in the total energy of the particle. That's very convenient, because then energy $E$ and momentum $\vec{p}$ together build a Minkowski four-vector which is manifestly Lorentz covariant. The invariant square of this four-vector is invariant under Lorentz transformations and given by
$$E^2/c^2-\vec{p}^2=m_0^2 c^2.$$
This shows that the mass of a particle is a scalar under Lorentz transformations and thus independent of the particle's velocity.

It's a very good practice in modern relativistic particle physics that we only use the notion of mass in the sense of "invariant mass", which is a Lorentz scalar. For a massive particle the invariant mass is sometimes also called the "rest mass" of the particle. For a massless particle, the invariant mass is zero, and the particle always moves with the speed of light.

For your problem you solve (1) for $p$ as a function of the invariant mass $m_0$ and the kinetic energy $|e V|$. This should lead you to the correct formula for the de Broglie wavelength.

If you are supposed to give the result in terms of the (three-)velocity of the particle, then you can use
$$p=\frac{m_0 v}{\sqrt{1-v^2/c^2}}.$$
It's easy to show that the energy of the particle (inlcuding its rest energy) is given by
$$E=\frac{m_0 c^2}{\sqrt{1-v^2/c^2}}.$$

Note: In the old days, before Minkowski had clarified the mathematical structure of relativistic spacetime (Minkowski space), the people used to introduce several kinds of "relativistic masses" that are speed dependent. Although unfortunately some modern textbooks still use these old-fashioned ideas, this is more confusing than helpful. So it's better to work only with the invariant mass, energy, and momentum.

Simen said:
Thank you so much for your reply, I'm still not entirely sure about how to og about this. I think the formula used for the energy in my post above is the formula for the kinetic energy, and the formula for the total energy would be:
Ek + mc2 = √(p^2 * c^2 + m^2*c^4)

which turns into:

Ek = √(p^2 * c^2 + m^2*c^4) - mc2
Ek 2 = p2 * c2 + 2m2*c4
Ek 2 / c2 = p2 +2m2 * c2
p2 = Ek 2 / c2 - 2m2 * c2

p = Ek / c - √(2m2 * c2)
or
p = √(eV2 / c2 -2m2 c2)

λ = h / √(eV2 / c2 -2m2 c2)

Is there a difference in the two masses? Should I put in a Lorentz factor somewhere to account for rest mass vs. mass at relativistic speed?
As vanhees noted, you should expunge the notion of relativistic mass from your brain. You also made some algebra errors. You seemed to have used ##(a+b)^2 = a^2+b^2## when squaring the expression for ##E_k## and ##\sqrt{a^2+b^2}=a+b## when solving for ##p##. Neither is correct.

1. What is the de Broglie wavelength of a particle?

The de Broglie wavelength of a particle is a concept in quantum mechanics that describes the wavelength of a particle, which is related to its momentum. It was first proposed by Louis de Broglie in 1924 and is given by the equation: λ = h/mv, where λ is the de Broglie wavelength, h is Planck's constant, m is the mass of the particle, and v is its velocity.

2. How is the de Broglie wavelength related to the particle's momentum?

The de Broglie wavelength is inversely proportional to the particle's momentum. This means that as the momentum increases, the wavelength decreases, and vice versa. This relationship is known as the de Broglie relation and is also used to describe the wave-particle duality of matter.

3. Can all particles have a de Broglie wavelength?

Yes, all particles, including massive particles like atoms and subatomic particles like electrons, have a de Broglie wavelength. However, the wavelength for macroscopic objects like humans or cars is incredibly small and therefore not observable.

4. What is the significance of the de Broglie wavelength?

The de Broglie wavelength is significant because it shows that particles, which were previously thought to only have particle-like properties, also have wave-like properties. This concept played a crucial role in the development of quantum mechanics and our understanding of the behavior of particles at the atomic and subatomic level.

5. How is the de Broglie wavelength experimentally observed?

The de Broglie wavelength is typically observed in experiments that involve diffraction or interference of particles, such as the famous double-slit experiment. In these experiments, the diffraction or interference patterns can only be explained by considering the de Broglie wavelength of the particles. Additionally, modern technologies like electron microscopes utilize the de Broglie wavelength to image objects at the nanoscale.

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