The K_ X-rays of aluminium ( Z = 13 ) and zinc ( Z = 30 ) possess wavelengths of 887 pm and 146 pm , respectively. A particular element emits a K_ X-ray with an energy of 3.69 keV . Using Moseley's law = a(Z - b) , identify this element.

Step-by-step solution

According to Moseley's law for K_ X-rays, the frequency is given by = a(Z - b) . For K_ lines, the screening constant b 1 . The energy of the emitted X-ray is given by: E = h = Rhc(Z-1)^2 ( 1 1^2 - 1 2^2 ) Substituting Rhc 13.6 eV : E 13.6 3 4 (Z-1)^2 eV = 10.2 (Z-1)^2 eV Given the energy of the unknown element's K_ X-ray is E = 3.69 keV = 3690 eV : 3690 = 10.2 (Z-1)^2 (Z-1)^2 = 3690 10.2 361.76 Z - 1 361.76 19.02 Z 20 The atomic number Z = 20 corresponds to Calcium. (Note: Using the given wavelengths for Al and Zn