A beam of protons and a beam of alpha particles (of mass 6.64 ×10^-27 kg and charge +2 e ) are a

A beam of protons and a beam of alpha particles (of mass...

A beam of protons and a beam of alpha particles (of mass $6.64 \times 10^{-27} \mathrm{kg}$ and charge $+2 e )$ are accelerated from rest through the same potential difference and pass through identical circular holes in a very thin, opaque film. When viewed far from the hole, the diffracted proton beam forms its first dark ring at $15^{\circ}$ with respect to its original dircction. When viewed similarly, at what angle will the alpha particle form its first dark ring?

Key Concepts

de Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics which states that particles exhibit wave-like properties. The wavelength is inversely proportional to the momentum of the particle, given by ? = h/p, where h is Planck’s constant. In the context of particles accelerated by a potential difference, their kinetic energy—and hence momentum—is determined by the work done by the electric field, which influences their de Broglie wavelength.

Acceleration Through a Potential Difference

When charged particles are accelerated from rest through a potential difference, they gain kinetic energy equal to the product of their charge and the potential difference (qV). This energy gain influences their velocity and momentum, and hence their de Broglie wavelength. The mass and charge of the particle dictate how much kinetic energy is acquired and how the momentum scales, which is critical when comparing particles like protons and alpha particles.

Diffraction from Circular Apertures

Diffraction of particles occurs when their de Broglie wavelengths are comparable to the dimensions of an aperture. When particles pass through a circular hole, they produce a diffraction pattern similar to light waves passing through a circular aperture. The angle to the first dark ring in the diffraction pattern (commonly determined by parameters related to the Bessel function) depends directly on the particle’s wavelength and the geometry of the aperture. Thus, any changes in the de Broglie wavelength, as due to differences in mass and charge of the particles, will alter the observed diffraction angles.

Transcript

00:01 In this exercise, we have a beam of protons that pass through a circular aperture, and the first diffraction minimum is obtained at the angle theta p of 15 degrees.

00:18 Then a second beam of particles, but this time of alpha particles, that have a mass of 6 .64 times 10 to the minus 27 kilograms, passes through the same circular aperture, so the diameter of the aperture is the same, and we want to know at what angle the first diffraction minimum is obtained.

00:57 First of all, i'm going to list here some constants that are going to be useful for us, so have the mass of the proton, that's 1 .67 times 10 to the minus 27 kilograms.

01:11 I feel like this is the constants that's going to be useful for us.

01:17 And also i'm going to write here, briefly remind you how to obtain the first diffraction minimum for a circular aperture.

01:25 So consider a circular aperture of diameter a, and the angle of the first diffraction minimum theta is equal to 1 .22 times the wavelength of the particle that's the going through the aperture.

01:50 So notice that for the proton, we have a times a sign of theta p is equal to 1 .22 times the wavelength of the proton.

02:04 While for the alpha particle, we have the a times the sign of theta alpha is equal to 1 .22 times the wavelength of the alpha particle.

02:16 So if we divide it, we divide both equations, we obtain that sign of theta p divided by the sine of theta alpha is equal to lambda p divided by lambda alpha.

02:41 And argois, you obtain theta alpha, so notice that sine of theta alpha is theta, lambda, divided by lambda p times the sign of theta p.

02:56 So theta alpha is the arc sine of lambda alpha divided by lambda p times the sign of theta p.

03:13 So notice that we already have theta p and we have to find theta alpha, i'm sorry, lambda alpha and lambda p in order to obtain theta alpha.

03:24 That is our ultimate goal.

03:27 So first of all, remember that for a general charged particle that accelerates from rest, the kinetic energy of the particle can be written as e times the potential difference delta v that accelerated that particle.

03:49 And another way to write the kinetic energy is the momentum squared divided by 2m...

Please give Ace some feedback