An X-ray tube works at 60 kV. Use hc = 1240 eV nm and h = 4.14 × 10⁻¹⁵ eV s…

Step 1. The shortest wavelength. The electron has $60000$ eV. So $\lambda_{\min} = \dfrac{1240\ \text{eV nm}}{60000\ \text{eV}} = 0.0207\ \text{nm} = 20.7\ \text{pm}$.

Step 2. The highest frequency. $f_{\max} = \dfrac{eV}{h} = \dfrac{60000\ \text{eV}}{4.14 \times 10^{-15}\ \text{eV s}} = 1.45 \times 10^{19}\ \text{Hz}$.

Step 3. The voltage for $10.0$ pm. $10.0\ \text{pm} = 0.0100$ nm. So $V = \dfrac{1240\ \text{eV nm}}{0.0100\ \text{nm}} \times \dfrac{1}{e} = 124000\ \text{V} = 124\ \text{kV}$.