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X-rays are a form of electromagnetic radiation with very short wavelengths. To give you perspective, they range from about 0.01 to 10 nanometers, which makes them much shorter than visible light. In physics, the wavelength is the distance between successive peaks of a wave. For x-rays, these waves oscillate extremely rapidly due to their short wavelengths. This property allows them to penetrate materials more deeply than visible light, making x-rays very useful in medical imaging and other applications. In the original exercise, the wavelength given is 35.0 picometers, or 35.0 pm. Converting this to meters for calculation, we have 35.0 pm = 35.0 × 10^{-12} m, illustrating the miniscule scale at which these waves operate.

Photon energy is directly related to its frequency through Planck's relation. The equation used is \( E = hf \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \). Frequency is inversely proportional to wavelength; hence, shorter wavelengths like x-rays lead to higher frequencies and consequently, more energetic photons.In practice:

• First, calculate the frequency \( f \) of the x-ray using \( f = \frac{c}{\lambda} \), identifying the speed of light \( c \) as \( 3 \times 10^8 \, \text{m/s} \).
• Next, substitute the frequency into the photon energy equation to find \( E \).
• Convert the result from joules to a more applicable unit like electron volts (eV), where 1 eV is the energy gained when an electron is accelerated through a potential difference of one volt.

This conversion gives you energy in kiloelectronvolts (keV), common in physics for high-energy phenomena like x-rays.

Momentum, in the realm of photon physics, might sound a bit abstract since photons are massless particles. However, they indeed carry momentum and this is pivotal in interactions like the Compton effect.For photons, momentum \( p \) is given by \( p = \frac{E}{c} \), where \( E \) is the energy calculated earlier, and \( c \) is the speed of light. It’s quite intriguing that even without mass, their momentum is significant due to their immense energy.To understand momentum better:

• The energy \( E \) is computed in kiloelectronvolts (keV) and should be converted back to joules for precise calculations.
• The photon momentum result can then be expressed as a unit of energy per speed of light, \( \text{keV}/c \), which aligns with quantum electrodynamics conventions.

This demonstrates how energy and momentum interplay even on the quantum scale, hinting at the wave-particle duality that defines the unique nature of light and other electromagnetic waves.