To find the wavelength ($\lambda$) of the shortest wavelength produced, we use the relationship between the speed of light ($c$), wavelength ($\lambda$), and frequency ($f$):

$c = \lambda f$

Where:

• $c$ is the speed of light ($3.00 \times 10^8 \, \text{m/s}$),
• $f$ is the frequency ($2.42 \times 10^{18} \, \text{Hz}$).

Rearranging the equation for $\lambda$:

$\lambda = \frac{c}{f}$

Substitute the known values:

$\lambda = \frac{3.00 \times 10^8 \, \text{m/s}}{2.42 \times 10^{18} \, \text{Hz}}$

Performing the calculation:

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

Final Answer:

The wavelength of the shortest wavelength produced is:

$\lambda = 1.24 \, \text{Å (angstrom)} \, \text{or} \, 0.124 \, \text{nm}.$

Let me know if you would like further clarification or detailed steps.

Questions to Explore Further:

1. What is the relationship between photon energy and wavelength?
2. How does the wavelength relate to the x-ray tube voltage (kVp)?
3. What role does frequency play in determining the energy of an x-ray photon?
4. How does this shortest wavelength compare with wavelengths in the visible spectrum?
5. What are practical applications of x-rays with such short wavelengths?

Tip:

Always convert units appropriately (meters, nanometers, angstroms) based on the context of the problem for clear interpretation.