A nucleus of the boron11 isotope consists of five protons and six neutrons. A particular ionized atom of boron11, whose mass is 1.831026kg,1.831026kg, lacks 33 electrons from its neutral state. Find the magnitude and direction of the electric field that will levitate this ion, exactly balancing its weight. Take =9.81m/s².

Answer rating: 100% (QA)

To find the magnitude and direction of the electric field that will levitate the boron11 ion we need ... View the full answer