(a) What is the energy of a photon that has wavelength 0.10μm ? (b) Through approximately what p

(a) What is the energy of a photon that has wavelength...

(a) What is the energy of a photon that has wavelength 0.10$\mu \mathrm{m} ?$ (b) Through approximately what potential difference must electrons be accelerated so that they will exhibit wave nature in passing through a pinhole 0.10$\mu \mathrm{m}$ in diameter? What is the speed of these electrons? (c) If protons rather than electrons were used, through what potential difference would protons have to be accelerated so they would exhibit wave nature in passing through this pinhole? What would be the speed of these protons?

Key Concepts

Mass Dependence in Wave-like Behavior

The mass of a particle significantly influences its speed for a given kinetic energy. Since electrons and protons have very different masses, for the same de Broglie wavelength or kinetic energy, their speeds will differ. This concept is crucial when comparing the acceleration requirements and resulting speeds for different particles, highlighting the role of mass in quantum mechanical behaviors.

Relationship Between Kinetic Energy, Momentum, and Speed

This involves linking a particle’s kinetic energy to its momentum and hence to its speed through the equations KE = ½ mv² and p = mv. Understanding this relationship is essential when determining how accelerating a particle to a certain kinetic energy affects its wave properties, as dictated by its de Broglie wavelength.

Photon Energy Calculation

This concept involves determining the energy of a photon using the relationship E = hc/?, where h is Planck’s constant and c is the speed of light. It is central to quantum mechanics as it links the particle-like properties of light with its wave characteristics, showing how wavelength is directly related to energy.

De Broglie Wavelength

The de Broglie wavelength concept posits that all particles exhibit wave-like properties, with their wavelength given by ? = h/p, where p is the momentum. This idea is fundamental to quantum mechanics and is used to predict phenomena such as electron diffraction when particles pass through small apertures.

Kinetic Energy from Accelerating Charged Particles

When charged particles like electrons or protons are accelerated through a potential difference, their kinetic energy is given by the equation KE = qV, where q is the charge and V is the potential difference. This allows one to connect the classical mechanics concept of energy gained to the quantum mechanical behavior via the de Broglie wavelength.

Transcript

00:02 Okay, so in this question we're talking about photon with a wavelength of 0 .1 micrometer.

00:08 So let's see where this problem goes.

00:12 E is equal to hc over lambda.

00:15 And then we were told lambda is 0 .1 micrometer, so it's 0 .1 times 10 to the minus 6 meters.

00:25 And what we want to do is do h times 0 .99 times 10.

00:32 To the eighth, and then divided by .1 times 10 to the minus 6.

00:41 Is that really what they asked for? what is the energy of a photon? and i guess i should put it in evv, so 1 .602 times 10 to the minus 19.

00:50 And so i divided it by e, the charge of the electron, and i get, how many sigfig should i do it to? two sig -fig.

00:58 So then i get 12b for a.

01:12 And then for b, we want to know that if we're accelerating electrons through some potential difference.

01:23 And we want them to exhibit their wave nature if they're going to pass through a pinhole of that size.

01:30 And then what's the speed of the electrons? oh, that's we want their, so basically we want their debroughly wave length to be on scale with the size of this pinhole.

01:41 So we, because in general, whenever the, whatever dimensions the debroughly wave length is, that's where you'll see the wave effects.

01:52 And so for normal matter, that becomes very, very small.

01:56 So we don't generally see the wave effects.

01:58 So what we're going to do then is do lambda, is use this for our lambda.

02:05 And then set that equal to h over p.

02:12 Okay, and then, okay, we want to know the potential difference.

02:25 Okay, so we have lambda, we have h.

02:27 So we can get p, so you can get that p is equal to h over lambda.

02:34 And then p.

02:37 So thinking about setting up, blah, blah, blah, let me back up.

02:40 So this is the p that we need to get.

02:42 So we can calculate that by just dividing this number by, or dividing age by this number.

02:48 But then to associate that with the voltage, we need to figure out how many, what potential difference would get an electron to that potential difference.

02:57 Or excuse me, to that momentum.

02:58 And so basically, through energy conservation, you can say you can turn the energy gain through being accelerated through potential difference, which is e delta v into kinetic energy, and that's p squared over 2m.

03:12 So then if you take this equation and solve for v, you just want to take this whole thing and divide it by e.

03:18 So delta v is equal to p squared over 2m.

03:24 I wonder if a will end up connecting.

03:27 We'll find out.

03:28 So p is plugging in this for p.

03:34 Then we get h squared over lambda squared.

03:41 M.

03:46 Oh, and then times two.

03:48 I can't forget this too...