CP (a) A particle with mass m has kinetic energy equal to...

$\mathrm{CP}$ (a) A particle with mass $m$ has kinetic energy equal to three times its rest energy.What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: $E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}$ and $\left.K=E-m c^{2} .\right)$ (b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

$\mathrm{CP}$ (a) A particle with mass $m$ has kinetic energy equal to three times its rest energy.What is the de Broglie wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: $E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}$ and $\left.K=E-m c^{2} .\right)$
(b) Determine the numerical value of the kinetic energy (in MeV) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (ii) a proton.

Key Concepts

Relativistic Energy-Momentum Relation

This relation, given by E² = (pc)² + (mc²)², is a cornerstone of special relativity. It connects a particle's total energy, momentum, and rest mass, allowing us to describe the energy and dynamics of particles moving at speeds comparable to the speed of light. This formula is essential for analyzing situations where classical mechanics fails to accurately represent the particle's behavior.

Relativistic Kinetic Energy

In special relativity, the kinetic energy is defined as the difference between the total energy and the rest energy (K = E - mc²). This definition takes into account the increased energy required to accelerate a particle as its velocity approaches the speed of light, distinguishing it from the classical expression for kinetic energy and ensuring accuracy in high-speed scenarios.

de Broglie Wavelength

The de Broglie wavelength concept posits that every particle exhibits wave-like behavior, with a wavelength given by ? = h/p, where h is Planck's constant and p is the particle's momentum. This idea, fundamental to quantum mechanics, underlies the wave-particle duality and is crucial for understanding phenomena such as diffraction and interference at the quantum scale.

Unit Conversions in Relativistic Calculations

Accurate calculations in relativistic physics often require converting between units, such as energy (e.g., joules to electron volts) and length (e.g., meters). Mastery of unit conversions is necessary to ensure consistency and correctness when applying theoretical formulas to practical problems, particularly when dealing with high-energy particles.

Transcript

00:01 Okay, here we have a tricky question where we're dealing with the kinetic energy and internal energy of a particle.

00:10 So we're told that some particle with mass m has kinetic energy, which is equal to three times its internal energy, and we're asked to find the deboile wavelength in question a.

00:26 So obviously we're going to use the deboile formula, which says that lambda is equals u .s.

00:31 Over.

00:31 Ic, or excuse me, mv.

00:37 But we need to find the velocity in order to value of lambda.

00:41 So let's write down as many equations as we know that relate to this.

00:46 So we have k equals through u.

00:49 We also know that k plus u has got to be equal to e.

00:54 That's just the definition.

00:58 And we know that e squared is equal to the quantity of pc squared plus the quantity of mc squared, quantity squared.

01:12 And then finally we know u is equal to mc squared.

01:17 So let's try to find something that relates to velocity from all these equations that we wrote down.

01:24 Pretty much all these are in terms of mc and p.

01:29 And we know that p is equal to nv.

01:34 So if we can find the momentum, we can divide that by m to get the velocity, which will plug into the way we're going on.

01:42 All right, so let's try to get everything in terms of m, c, and p.

01:48 So first, let's isolate k because k is something that doesn't immediately have an easy substitution.

01:58 We know what e is if we just square root this and we know what u is.

02:01 So let's get everything else in terms of k.

02:05 So first, k is going to be equal to e minus u from this equation, which is also going to be equal to three u from this equation.

02:16 All right, so now let's sub in what we know e and u are, and then we can solve for p, and then consequently v, and we'll plug that into the toy of the formula.

02:26 All right, so let's sub everything in.

02:28 So e is going to be the square root of quantity, pc quantity squared plus quantity of mc squared, quantity squared, quantity squared...

Please give Ace some feedback