An ion that is doubly ionized passes through the velocity selector and into the deflection chamber of a mass spectrometer, as shown below. In the velocity selector the electric field has a magnitude of 8464 V/m, and the magnitude of the magnetic field in both the velocity selector and the deflection chamber is 0.0943 T. If in the deflection chamber the ion is detected at a distance of 11.9 cm from its entry point, determine the following.(a) mass-to-charge ratio of the ionkg/C(b) mass of the ionkg(c) identity of the ion, assuming it's an element (Use only the masses of elements in their most common form as listed on the periodic table of elements.) **Diagram Explanation:**This illustration depicts an electric and magnetic field interaction on a charged particle. Here's a detailed breakdown of the components:1. **Charged Particle:** - A positively charged particle is indicated at the bottom, denoted by a "+". It is initially moving vertically upward with velocity \(\vec{v}\).2. **Electric Field (\(\vec{E}\)):** - Represented by an orange arrow pointing to the left, indicating the direction of the electric field. This field exerts force on the charged particle perpendicular to its initial velocity.3. **Magnetic Field (\(\vec{B}\)):** - Denoted by blue circles with dots at the center, indicating that the magnetic field is pointing out of the plane of the page.4. **Forces and Motion:** - The force due to the electric field will act to the left, redirecting the movement of the charged particle horizontally. - As the particle enters the region with the magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field direction (perpendicular to the page). This results in a circular motion of the particle, with the radius denoted by \(r\). - The dashed line shows the path of the charged particle under the combined influence of the electric and magnetic fields.5. **Conducting Plates:** - Two conducting plates are on either side of the particle's initial path. The left plate is marked with negative signs (-), and the right plate with positive signs (+). These are likely the source of the electric field \(\vec{E}\).This diagram effectively illustrates the interaction and resultant motion when a charged particle is subjected to perpendicular electric and magnetic fields.

Given Data: The magnitude of an electric field is, E=8464 V/m The magnitude of the magnetic field is, B=0.0943 T The radius of motion is calculated as, r=11.9 cm2r=5.95 cm×10-2 m1 cmr=0.0595 m Part (a): The expression for the electric force is given as, FE=qE...............(1) Here, q is the charge and E is the electric field. The expression for the magnetic force is given as, FB=Bqv..............(2) Here, B is the magnetic field and v is the speed. From equation (1) and (2), electric force balances the magnetic force. So, FE=FBqE=Bqvv=EBv=8464 V/m0.0943 Tv=89.75×103 m/s Now, the centripetal force is provided by the magnetic force then, mv2r=Bqvmq=Brvmq=0.0943 T×0 .0595 m89.75×103 m/smq=6.25×10-8 kg/C Therefore, the mass-to-charge ratio of the ion is 6.25×10-8 kg/C.Part (b): It is given that the ion is doubly ionized then its charge will be, Q=2×1.6×10-19 CQ=3.2×10-19 C The mass of the ion is, m=3.2×10-19 C×6.25×10-8 kg/Cm=2×10-26 kg Therefore, the mass of the ion is 2×10-26 kg. Part (c): The mass of one nucleon is 1.6×10-27 kg. Hence, the number of nucleons is calculated as, N=2×10-26 kg1.6×10-27 kgN=12.5 ∴N≈13N=13 Hence, the aluminium ion is doubly ionized.