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The frequency of a photon is a fundamental concept in understanding electromagnetic waves like X-rays. Frequency, denoted as \( u \), describes how often the wave oscillates per second. It's measured in Hertz (Hz). For any electromagnetic wave, the frequency is inversely proportional to its wavelength, represented by the formula: \[ u = \frac{c}{\lambda} \] Here, \( c \) is the speed of light, which is a constant \( 3 \times 10^8 \text{ m/s} \), and \( \lambda \) is the wavelength. In our problem, we converted the X-ray wavelength from picometers to meters, which allowed us to find the frequency.

• A shorter wavelength means a higher frequency, and vice versa.
• For a wavelength of \( 35.0 \text{ pm} \), which is the same as \( 35.0 \times 10^{-12} \text{ m} \), we calculated the frequency to be approximately \( 8.57 \times 10^{18} \text{ Hz} \).

Understanding frequency is key because it directly affects other properties of the photon like energy and momentum.

Photon energy is the amount of energy carried by a single photon, and it can be calculated using the formula: \[ E = h u \] Here, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \text{ Js} \), and \( u \) is the frequency. Photons of higher frequency X-rays inherently carry more energy. In our exercise, we first calculated the frequency, which allowed us to then determine the energy of the photon.

• Using \( u = 8.57 \times 10^{18} \text{ Hz} \), the energy \( E \) was found to be \( 5.68 \times 10^{-15} \text{ J} \).
• This value can convert to more common units for photon energies, kiloelectron volts (keV), resulting in approximately \( 35.46 \text{ keV} \).

Understanding photon energy helps us appreciate many phenomena in physics, such as how X-rays can penetrate materials and affect imaging.

Photon momentum might seem counterintuitive since photons are massless, but they indeed possess momentum, which can be described using the energy-momentum relationship: \[ p = \frac{E}{c} \] In this formula, \( E \) is the photon energy (in joules), and \( c \) is the speed of light. As previously noted, photon energy is derived from its frequency, having already been calculated in joules before conversion to electron volts.

• With \( E = 5.68 \times 10^{-15} \text{ J} \), the momentum \( p \) was computed to be about \( 1.89 \times 10^{-23} \text{ kg m/s} \).
• Converting into the unit keV/c, which streamlines comparison and applications in particle physics, this came out as approximately \( 3.54 \text{ keV/c} \).

The concept of photon momentum is vitally important in advanced physics, allowing us to understand interactions such as light pressure and momentum transfer in collisions.