To find the frequency of an x-ray photon with an energy of 600 keV, we use the relationship between energy and frequency in quantum mechanics:

$E = h \cdot f$

Where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant ($6.626 \times 10^{-34} \, \text{J·s}$),
• $f$ is the frequency of the photon.

Since the energy is given in keV (kiloelectron volts), we need to convert it to joules. The conversion factor is:

$1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$

Thus, 600 keV is:

$600 \, \text{keV} = 600 \times 10^3 \, \text{eV} = 600 \times 10^3 \times 1.602 \times 10^{-19} \, \text{J} = 9.612 \times 10^{-14} \, \text{J}$

Now we can solve for the frequency:

$f = \frac{E}{h} = \frac{9.612 \times 10^{-14}}{6.626 \times 10^{-34}} \, \text{Hz}$

Let me calculate that for you.The frequency of the x-ray photon with an energy of 600 keV is approximately $1.45 \times 10^{20}$ Hz.

Would you like further details on this calculation or have any questions?

Here are some related questions you might find interesting:

1. What is the wavelength of a photon with a frequency of $1.45 \times 10^{20}$ Hz?
2. How is the energy of a photon related to its wavelength?
3. What is the energy of a visible light photon compared to an x-ray photon?
4. How does the energy of photons affect their interaction with matter?
5. How does the photoelectric effect relate to photon energy?

Tip: The higher the frequency of a photon, the higher its energy and ability to penetrate matter!