Question

In a TV picture tube the accelerating voltage is 17.0 kV, and the electron beam passes through an aperture 0.40 mm in diameter to a screen 0.230 m away. What is the uncertainty in position of the point where the electrons strike the screen?

Answer 1

Each electron has a charge of e=1.60217657 × 10^-19 C, and the accelerating voltage (which is actually the potential difference between the filament and the anode in the TV) is 17000 volts.

The electric energy that is transferred to the electrons can be expressed as

eV = 1.602 × 10^-19 C x 17000 V= 2.723 × 10^-15 J (note that the product C × V= J, the SI unit of energy)

Then, when the electrons travel through the TV tube, they are accelerated, so they gain kinetic energy, and it can be expressed as E_k=1/2 m_ev²

Where m_e=9.109 × 10^-31 kg is the mass of an electron, and v is the speed, in m/s.

All the electric energy transferred to the electrons changes into kinetic energy: eV=1/2 m_ev²

Solving for v, we can find the speed of the electrons:

Then, we determine the wavelength of the electron beam using the DeBroglie equation (h is the Planck constant):

After the beam passes throught the aperture, an angular separation occurs. We can find the vertical distance y from the centerline of the beam at which the electrons strike the beam as

where m is an integer related to the order of the diffracted beam, D is the distance between the aperture and the screen, and a is the diameter of the aperture.

Then, the uncertainty in position of the point where the electrons strike the screen would be twice the distance y from the center of the beam, so .

Note that the question is not about the uncertainty in the position along the x axis, that we could have found using Heisenberg's uncertainty principle, because the distance between the aperture and the screen is already known (the electrons strike the screen after they travel 0.230 m from the aperture).