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**Example:** Find the wavelength of X‑rays produced when electrons are accelerated through a potential difference of 30 kV. **Step 1:** Convert the voltage to electron‑volts (eV). 1 kV = 1000 eV, so 30 kV = 30,000 eV. **Step 2:** Use the relation between photon energy and wavelength: \(E = \frac{hc}{\lambda}\). **Step 3:** Rearrange to get \(\lambda = \frac{hc}{E}\). **Step 4:** Insert the constants: \(h = 6.626\times10^{-34}\,\text{J·s}\), \(c = 3.00\times10^{8}\,\text{m/s}\), and convert energy to joules: \(E = 30,000\,\text{eV} \times 1.602\times10^{-19}\,\text{J/eV}=4.806\times10^{-15}\,\text{J}\). **Step 5:** Calculate: \[\lambda = \frac{6.626\times10^{-34}\times3.00\times10^{8}}{4.806\times10^{-15}} \approx 4.13\times10^{-11}\,\text{m}\] **Result:** The X‑ray wavelength is about 0.041 nm (or 41 pm). This short wavelength explains why X‑rays can penetrate materials and produce fine details in radiographs.