The electrons in a cathode-ray tube are accelerated through a 250 V potential difference and then shot through a 33-nm-diameter circular aperture. What is the diameter of the bright spot on an electron detector 1.5 m behind the aperture?

Verified step by step guidance

1. Determine the speed of the electrons after being accelerated through the 250 V potential difference. Use the energy conservation principle: the kinetic energy gained by the electrons is equal to the work done by the electric field. The formula is \( \frac{1}{2}mv^2 = eV \), where \( m \) is the mass of the electron, \( v \) is its velocity, \( e \) is the charge of the electron, and \( V \) is the potential difference.
2. Calculate the de Broglie wavelength of the electrons using the formula \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( v \) is the velocity obtained in the previous step.
3. Determine the angular width of the central diffraction maximum using the single-slit diffraction formula for electrons: \( \sin \theta = \frac{1.22 \lambda}{D} \), where \( \lambda \) is the de Broglie wavelength, and \( D \) is the diameter of the aperture (33 nm).
4. Calculate the linear diameter of the bright spot on the detector. The formula is \( d = 2L \tan \theta \), where \( L \) is the distance from the aperture to the detector (1.5 m), and \( \tan \theta \approx \sin \theta \) for small angles.
5. Substitute all known values into the equations and simplify to find the diameter of the bright spot. Ensure all units are consistent (e.g., convert nm to meters) before performing calculations.

Key Concepts

Electron Acceleration

When electrons are subjected to a potential difference, they gain kinetic energy proportional to the voltage. In this case, a 250 V potential difference accelerates the electrons, allowing us to calculate their final velocity using the equation KE = eV, where KE is kinetic energy, e is the charge of the electron, and V is the voltage.

Wave-Particle Duality

Electrons exhibit both particle-like and wave-like properties, a concept known as wave-particle duality. When electrons pass through the aperture, they can be treated as waves, which leads to diffraction. This behavior is crucial for determining the size of the bright spot on the detector, as it depends on the wavelength of the electrons.

Diffraction and Spot Size

Diffraction occurs when waves encounter an obstacle or aperture, causing them to spread out. The size of the bright spot on the detector can be calculated using the diffraction formula, which relates the aperture size, distance to the detector, and the wavelength of the electrons. This relationship helps predict how the wave nature of electrons affects the observed spot size.