A proton, a deuteron (q=+e, m=2.0 u), and an alpha particle (q=+2 e, m=4.0 u) are accelerated th

A proton, a deuteron (q=+e, m=2.0 u), and an alpha particle...

A proton, a deuteron $(q=+e, m=2.0 \mathrm{u})$, and an alpha particle $(q=+2 e, m=4.0 \mathrm{u})$ are accelerated through the same potential difference and then enter the same region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. What is the ratio of (a) the proton's kinetic energy $K_{p}$ to the alpha particle's kinetic energy $K_{a}$ and (b) the deuteron's kinetic energy $K_{d}$ to $K_{n} ?$ If the radius of the proton's circular path is $10 \mathrm{~cm}$, what is the radius of (c) the deuteron's path and (d) the alpha particle's path?

A proton, a deuteron $(q=+e, m=2.0 \mathrm{u})$, and an alpha particle $(q=+2 e, m=4.0 \mathrm{u})$ are accelerated through the same potential difference and then enter the same region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. What is the ratio of (a) the proton's kinetic energy $K_{p}$ to the alpha particle's kinetic energy $K_{a}$ and (b) the deuteron's kinetic energy $K_{d}$ to $K_{n} ?$ If the radius of the proton's circular path is $10 \mathrm{~cm}$, what is the radius of
(c) the deuteron's path and (d) the alpha particle's path?

Key Concepts

Acceleration of Charged Particles by a Potential Difference

When a charged particle is accelerated through a potential difference (?V), it gains kinetic energy equal to the product of its charge and the potential difference, expressed as K = q?V. This principle means that for particles experiencing the same ?V, the kinetic energy they gain is directly proportional to their charge, regardless of their mass.

Kinetic Energy and Particle Velocity

The kinetic energy (K) of a particle is related to its mass (m) and velocity (v) through the expression K = ½mv². This relationship is used to determine the particle’s speed after acceleration given its kinetic energy, and is a fundamental concept in classical mechanics for relating energy to motion.

Motion in a Uniform Magnetic Field

A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic (Lorentz) force that acts perpendicular to both its velocity and the field. This force causes the particle to move in a circular path, and by equating the magnetic force (qvB) to the centripetal force required for circular motion (mv²/r), the radius of the path can be derived as r = mv/(qB). This equation underscores the importance of the charge-to-mass ratio and the particle’s velocity in determining the curvature of its trajectory.

Charge?to?Mass Ratio Effects

The charge-to-mass ratio (q/m) plays a crucial role in both the energy gain during acceleration and the dynamics of motion in a magnetic field. A higher charge or a lower mass results in a higher kinetic energy for a given potential difference and also leads to a smaller radius of curvature in a magnetic field. Understanding this ratio is essential for comparing the behaviors of different charged particles under similar conditions.

Transcript

00:02 Here we're going to be looking at different positively charged particles and comparing the circular orbits that they make once they enter the same magnetic field after being accelerated by the same potential difference.

00:28 So that is our goal.

00:32 And so the first step is conservation of energy.

00:36 We want to compare their kinetic energies.

00:40 Do they all have the same kinetic energy as they get accelerated? so we're looking at the region of potential acceleration.

00:58 Okay, so we have their electrical potential energy converted into their kinetic energy as they enter the field, the magnetic field.

01:10 Electric potential energy is q times delta v.

01:20 And so our three charges are not going to have the same kinetic energy.

01:25 We can take a look.

01:27 But the kinetic energy of the proton compared to the alpha, for example, let's do the deuteron.

01:39 Let's do each pair.

01:41 Proton to deuteron, same charge.

01:47 So that would be e delta v over e delta v or 1.

01:57 The proton compared to the alpha particle, that's pretty simple to do.

02:06 E delta v over 2e delta v is equal to one half.

02:15 And then of course the duteron to the alpha particle is also going to be one half.

02:28 Yeah, so that's pretty simple.

02:29 And why we need that, we'll see that.

02:33 So as they enter the field, what we have here is there is a circular motion.

02:42 So by newton's second law, the centripetal force on each of the charges is mv squared over the radius.

02:52 And we will use magnetic force for the centripetal force.

02:57 Causing the curvature.

03:02 And we get that qvb equals mv squared over r, or solving for the r, which is what we're after, is mv over qb.

03:25 And it would help us if we wrote mv as the momentum.

03:34 Since we've already compared momenta, sorry, kinetic energies, we want to work out momentum in terms of kinetic energy.

03:45 P squared over 2m is equal to kinetic energy, or p is the square root of 2mk.

04:00 Okay, and so finally working this out, our final relationship is the radius goes as the square root of two the mass of the particle times its kinetic energy over its charge in the magnetic field...

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