A certain element emits K base α X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.

E = 3.69 kev = 3690 eV λ hc/E = 1242/3690 = 0.33658 nm √c/λ = a(z - b); a = 5 * 10^7 √Hz , b = 1.37 (from previous problem) √3 * 10^8/0.34 * 10^9 = 5 * 10^7 (Z – 1.37) ⇒ √8.82 * 10617 = 5 * 10^7 (Z – 1.37) ⇒ 9.39 * 10^8 = 5 * 10^7 (Z – 1.37) ⇒ 93.9 / 5 = Z – 1.37 ⇒ Z = 20.15 = 20 ∴ the element is calcium.