To find the minimum wavelength of the X-rays produced, we can use the relationship between the energy of the X-ray photons and their wavelength. The energy of a photon $E$ is related to its wavelength $\lambda$ by the equation:

$E = \frac{hc}{\lambda}$

where:

• $h$ is Planck's constant, $h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s}$,
• $c$ is the speed of light, $c = 3.00 \times 10^8 \, \text{m/s}$,
• $\lambda$ is the wavelength of the photon in meters.

The energy $E$ of the X-ray photon produced by the machine is equal to the energy associated with the maximum voltage applied, $V = 100 \, \text{kV}$. This energy can be calculated using the electron charge $e$ as:

$E = eV$

where $e = 1.602 \times 10^{-19} \, \text{C}$ and $V = 100 \, \text{kV} = 100,000 \, \text{V}$.

1. Calculate the energy $E$:

   $E = (1.602 \times 10^{-19} \, \text{C})(100,000 \, \text{V}) = 1.602 \times 10^{-14} \, \text{J}$

2. Rearrange the photon energy equation to solve for $\lambda$:

   $\lambda = \frac{hc}{E}$

3. Substitute the values:

   $\lambda = \frac{(6.626 \times 10^{-34} \, \text{J} \cdot \text{s})(3.00 \times 10^8 \, \text{m/s})}{1.602 \times 10^{-14} \, \text{J}}$

4. Calculate $\lambda$:

   $\lambda = 1.24 \times 10^{-11} \, \text{m} = 0.0124 \, \text{nm}$

Answer: The minimum wavelength of the X-rays produced by the machine is approximately $0.0124 \, \text{nm}$.

Would you like further details on the derivation or any step clarified?

---

Here are five related questions:

1. What happens to the minimum wavelength if the voltage is increased?
2. How does Planck's constant influence the wavelength of electromagnetic radiation?
3. What is the relationship between photon energy and frequency?
4. How can we calculate the energy of X-rays in electron volts (eV)?
5. What are the applications of knowing the minimum wavelength in X-ray imaging?

Tip: Remember, shorter wavelengths mean higher energy X-rays, which can penetrate materials more effectively.