Problem 33 - Fundamentals of Physics

Question

Through what minimum potential difference must an electron in an x-ray tube be accelerated so that it can produce \(x\) rays with a wavelength of \(0.100 \mathrm{~nm}\) ?

Step-by-Step Solution

Answer

12,400 volts

Step 1: Understanding the Problem

We need to find the minimum potential difference through which an electron must be accelerated to produce X-rays with a given wavelength of \(0.100 \mathrm{~nm}\). This involves concepts from quantum physics and the relationship between energy and wavelength.

Step 2: Using the Planck-Einstein Relation

The energy \(E\) of a photon is given by the Planck-Einstein relation: \(E = \frac{hc}{\lambda}\), where \(h\) is the Planck's constant \(6.626 \times 10^{-34} \text{ J s}\), \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength \(0.100 \mathrm{~nm} = 0.100 \times 10^{-9} \text{ m}\). Substitute these values in to find \(E\).

Step 3: Calculating the Energy

Substitute the constants into the energy formula: \(E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{0.100 \times 10^{-9}}\). This simplifies to approximately \(1.986 \times 10^{-15} \text{ J}\).

Step 4: Converting Energy to Electronvolts

The energy in joules needs to be converted to electronvolts (eV) where \(1 \, \text{eV} = 1.602 \times 10^{-19} \text{ J}\). Use the conversion: \(E_{ev} = \frac{E_{joules}}{1.602 \times 10^{-19}}\), which gives approximately \(12,400 \, \text{eV}\).

Step 5: Relating Energy to Potential Difference

The relationship between energy and potential difference is \(E = qV\), where \(q\) is the charge of an electron \(1 \, ext{eV}\ per electron\). Thus, \(V = E\) in eV, meaning \(V = 12,400 \, \text{volts}\). The minimum potential difference required is 12,400 volts.

Key Concepts

Planck-Einstein relation

The Planck-Einstein relation forms the backbone of our understanding of the energy of photons in quantum physics. Simply put, this relation connects the energy of a photon (a tiny particle of light) to its wavelength. The formula is given as \( E = \frac{hc}{\lambda} \), where \( E \) is the energy, \( h \) is Planck's constant \(6.626 \times 10^{-34} \mathrm{~Js} \), \( c \) is the speed of light \(3 \times 10^8 \mathrm{~m/s}\), and \( \lambda \) is the wavelength of light.

This relation beautifully ties together the wave and particle nature of light. Photons have energy, even though they are massless, and this energy is inversely proportional to the wavelength.

In practical terms, this means that shorter wavelengths (like X-rays) have more energy compared to longer wavelengths (like radio waves). Understanding this connection is crucial for applications ranging from medical imaging to quantum technology.

Wavelength and Energy Relationship

Photon energy and wavelength are directly related by the equation in the Planck-Einstein relation. A shorter wavelength means higher energy. This relationship is vital because it helps determine the energy levels involved in processes like the creation of X-rays.

To solve problems involving these concepts, it’s essential to convert the wavelength from nanometers (typically used for X-rays) to meters before substiting into the energy equation:\[ E = \frac{hc}{\lambda} \].

This conversion ensures consistency with SI units and allows for calculating energy in joules. Once we have the energy in joules, we can convert it to electronvolts (a more common unit in physics) using the conversion factor \(1 \, \text{eV} = 1.602 \times 10^{-19} \text{ J}\), which is immensely useful in many scientific calculations.

Electron Acceleration Potential

Electrons are fundamental particles with negative charge that can be accelerated through a potential difference (voltage). This process is crucial in generating X-rays. The potential difference needed for particle acceleration is directly related to the energy those electrons will obtain.

In an X-ray tube, electrons are accelerated to high speeds, acquiring kinetic energy which is equivalent to the difference in electrical potential energy expressed as \( E = qV \). Here, \( q \) is the charge of an electron, and \( V \) is the potential difference in volts.

Effectively, this means if you know the energy a particle needs (expressed in electronvolts), it also tells you the necessary potential difference. For instance, producing X-rays with a desired wavelength involves calculating the minimum potential difference required for electrons. The energy of the X-ray photon dictates this potential, which for the current problem is found to be \( 12,400 \text{ volts}\). Understanding this principle allows for better control over particle accelerators and similar technologies.