In a TV picture tube the accelerating voltage is 15.0 kV , and the electron beam passes through

In a TV picture tube the accelerating voltage is 15.0 kV ,...

In a TV picture tube the accelerating voltage is 15.0 $\mathrm{kV}$ , and the electron beam passes through an aperture 0.50 $\mathrm{mm}$ in diameter to a screen 0.300 $\mathrm{m}$ away. (a) Calculate the uncertainty in the component of the electron's velocity perpendicular to the line between aperture and screen. (b) What is the uncertainty in position of the point where the electrons strike the screen? (c) Does this uncertainty affect the clarity of the picture significantly? (Use nonrelativistic expressions for the motion of the electrons. This is fairly accurate and is certainly adequate for obtaining an estimate of uncertainty effects.)

Key Concepts

Heisenberg Uncertainty Principle

This fundamental principle of quantum mechanics states that the product of the uncertainties in position and momentum of a particle cannot be arbitrarily small, but is bounded by a value on the order of Planck’s constant. It provides the theoretical underpinning for estimating the minimum uncertainty in an electron's velocity when its position is confined by an aperture, demonstrating that a precise position measurement leads to a corresponding uncertainty in momentum.

Electron Acceleration and Kinetic Energy

In the context of accelerated electrons, the nonrelativistic kinetic energy acquired from an applied electric potential is given by the energy conversion from electric potential energy to kinetic energy. This relationship, expressed as eV = (1/2)mv², is used to calculate the electron’s velocity after acceleration, which then plays a key role in determining the electron’s momentum and its related uncertainties.

Diffraction through an Aperture

When waves, including the matter waves associated with electrons, pass through a narrow aperture, they undergo diffraction, which causes a spread in the wave’s momentum distribution perpendicular to the direction of travel. This diffraction effect is a physical manifestation of the uncertainty principle, where the confined spatial extent (the aperture size) leads to an inherent uncertainty in the momentum and, consequently, the velocity component transverse to the propagation direction.

Transcript

00:01 For this question, we have our electron passing through an aperture and getting diffracted to a screen.

00:13 A screen some distance 0 .3 meters away.

00:19 And as it passes through this slit or this aperture, its uncertainty in its position is confined to the size of this aperture.

00:31 And this is by the uncertainty principle this will affect the minimum uncertainty in the momentum in this particular direction which we will call the y -axis is the y -d -saxis so in the white axis the uncertainty in the position is fixed to be 0 .5 millimeters and this will affect the uncertainty in the momentum, which will be greater or equals to, which part of a 2 times 0 .5, stand by minus 3.

01:23 And if you will to calculate the momentum, we can actually calculate what is the uncertainty in the velocity by dividing by m.

01:37 So we take delta p further by m, where m is the mass of the electron.

01:54 So there's just 9 .11 times 10 power minus 31 should get an uncertainty of about 0 .12 meters per second.

02:15 Now to find the uncertainty in the position of the point when it strikes this screen, say we have an electron that is moving towards the screen.

02:30 Now because of the uncertainty in the wide direction, right, in the velocity in the wide direction, right, in the worst case scenario, we have some electron that is moving up or down with a velocity of 0 .12 meters per second.

02:52 This is the worst case scenario where there's a spread of velocity in 0 .12 meters per second, either up or down.

03:07 The further away it travels, the longer it will actually affect our distance when it reaches the screen.

03:20 So let's say if it travels for a short distance, then the uncertainty in the point would just be equal to 0 .12 meters per second, multiply by the time taken to travel this place.

03:38 Then this will be the maximum uncertainty in the point before it reaches that place.

03:54 Idea, right? when we travel further towards the screen, there is this uncertainty in the velocity.

04:02 So in the worst case scenario, if we have a particle that is traveling with an extra velocity downwards of 0 .12, we want to find what is the position it will reach...

Please give Ace some feedback