A proton (charge +e, mass mp ), a deuteron (charge +e, mass...

A proton (charge $+e,$ mass $m_{p} ),$ a deuteron (charge $+e$, mass $2 m_{p} ),$ and an alpha particle (charge $+2 e,$ mass 4$m_{p} )$ are accelerated from rest through a common potential difference $\Delta V$ . Each of the particles enters a uniform magnetic field $\overrightarrow{\mathbf{B}},$ with its velocity in a direction perpendicular to $\overrightarrow{\mathbf{B}}$. The proton moves in a circular path of radius $r_{p}$ . In terms of $r_{p},$ determine (a) the radius $r_{d}$ of the circular orbit for the deuteron and (b) the radius $r_{\alpha}$ for the alpha particle.

Key Concepts

Circular Motion in a Magnetic Field

A charged particle moving perpendicular to a uniform magnetic field experiences a magnetic Lorentz force that acts as the centripetal force, causing the particle to move in a circular path. The balance of the magnetic force (qvB) and the centripetal force (mv²/r) leads to the expression for the radius of curvature of the path (r = mv/(qB)). This concept is key in understanding particle trajectories in magnetic spectrometers and similar devices.

Substitution of Kinetic Energy into the Radius Formula

By equating the kinetic energy gained (q?V) to (1/2)mv², one can solve for the speed (v = ?(2q?V/m)). This speed, when substituted into the radius formula for circular motion in a magnetic field, gives an expression for the radius in terms of fundamental quantities (r = ?(2m?V)/(qB)). This derivation ties together energy conservation and force balance and is pivotal in determining the trajectories of different charged particles under the same experimental conditions.

Kinetic Energy Gain in an Electric Field

When a charged particle is accelerated from rest through a potential difference, it gains kinetic energy equal to the product of its charge and the voltage (q?V). This principle is essential because it allows us to relate the particle’s mass and charge to its final speed after acceleration, an idea that is broadly applied in fields such as accelerator physics.

Transcript

00:01 All righty, so we've got a special problem here today.

00:03 We've got three different particles that were investigating.

00:07 We've got a proton.

00:12 Proton.

00:13 Okay? this guy's got a charge of plus e, mass of m subpe.

00:23 We also have the deterrent.

00:31 And of course, this guy has the plus e charge and twice the mass of the proton.

00:39 And of course, we got the alpha particle.

00:42 Alpha particle.

00:45 This guy's interesting.

00:47 It's got twice the charge four times the mass of the proton, or twice the mass of a deterrent.

00:55 I'm going to box these in, and we're going to hold on to them.

00:59 All right.

01:00 So let's get to work here.

01:02 You ultimately want to figure out a way to have the radius of the deterron and the alpha particle in terms of the radius of the radius of the proton, given a situation where you have a potential difference, delta v, and you have a magnetic field perpendicular in motion.

01:21 All right, let's dive in.

01:24 First thing i want to start with here is going to apply conservation of energy since my system is an isolated system for energy.

01:35 Go ahead, that's going to be delta k plus delta u.

01:38 Those guys are equal to zero.

01:40 I'm going to go ahead and plug in some information we know here.

01:42 For the delta k, we have k final minus k initial plus the, and delta you, we're going to replace that with a negative q delta.

01:51 By definition q delta v still equal to zero now remember question states that the particles start from rest if they start from rest this kinetic energy term is there's nothing there's no kinetic energy initially because there's no velocity and velocity is a key component of kinetic energy a little bit more rearranging kinetic energy final equals q delta v and we'll take this one step forward more the one half m the final squared equals q delta b.

02:28 All right.

02:28 Now we're going to go ahead and solve this for velocity.

02:35 And rearranging, that's going to be 2 q delta v divided by mass.

02:42 And there's still that radical here because the v is square here.

02:46 So we have to get that radical.

02:47 All right.

02:48 So that's your standard velocity right there for your situation...

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