b) Singly charged ions of \( K_{19}^{39} \) and \( K_{19}^{41} \) injected into the mass spectrometer with a magnetic flux density \( \mathrm{B}=0.70 \mathrm{~T} \). (i) What is the speed \( v_{i} \) of the ions injected into the spectrometer? (ii)Determine the speed at which the ions hit the film, \( v_{f} \). (iii)Calculate the distance OP for each ion.

To solve the problem, we will use the principles of mass spectrometry, specifically the motion of charged particles in a magnetic field. ### Given Data: - Mass of \( K_{19}^{39} \) (Potassium-39) = 39 u - Mass of \( K_{19}^{41} \) (Potassium-41) = 41 u - Charge of the ions \( q = +1e \) (where \( e \approx 1.6 \times 10^{-19} \, \text{C} \)) - Magnetic flux density \( B = 0.70 \, \text{T} \) ### (i) Speed \( v_i \) of the ions injected into the spectrometer In a mass spectrometer, the speed of the ions can be determined using the relationship between the magnetic force and the centripetal force acting on the ions. The magnetic force \( F_B \) acting on a charged particle moving in a magnetic field is given by: \[ F_B = qvB \] The centripetal force \( F_c \) required to keep the ion moving in a circular path is given by: \[ F_c = \frac{mv^2}{r} \] Setting these two forces equal gives: \[ qvB = \frac{mv^2}{r} \] Rearranging this equation to solve for the speed \( v \): \[ v = \frac{qBr}{m} \] ### (ii) Speed at which the ions hit the film \( v_f \) The speed at which the ions hit the film is the same as the speed at which they were injected into the spectrometer, assuming no energy loss. Therefore: \[ v_f = v_i \] ### (iii) Calculate the distance OP for each ion The radius \( r \) of the circular path of the ions in the magnetic field can be derived from the equation: \[ r = \frac{mv}{qB} \] Substituting \( v \) from the earlier equation: \[ r = \frac{m}{qB} \cdot \frac{qBr}{m} = \frac{mv}{qB} \] This means that the distance OP (the radius of the circular path) can be calculated for each ion using their respective masses. ### Calculations 1. **For \( K_{19}^{39} \)**: - Mass \( m_{39} = 39 \, \text{u} = 39 \times 1.66 \times 10^{-27} \, \text{kg} \approx 6.48 \times 10^{-26} \, \text{kg} \) - Charge \( q = 1.6 \times 10^{-19} \, \text{C} \) - Magnetic field \( B = 0.70 \, \text{T} \) Using the formula for speed: \[ v_{39} = \frac{qBr}{m_{39}} \] To find \( r \): \[ r_{39} = \frac{m_{39}v_{39}}{qB} \] 2. **For \( K_{19}^{41} \)**: - Mass \( m_{41} = 41 \, \text{u} = 41 \times 1.66 \times 10^{-27} \, \text{kg} \approx 6.81 \times 10^{-26} \, \text{kg} \) Similarly, we can calculate: \[ v_{41} = \frac{qBr}{m_{41}} \] And for the radius: \[ r_{41} = \frac{m_{41}v_{41}}{qB} \] ### Summary of Results 1. **Speed \( v_i \) and \( v_f \)**: The speed of the ions is the same as they are injected and hit the film. 2. **Distance OP**: The radius of the circular path for each ion can be calculated using the mass of the ions and the magnetic field. To finalize the calculations, we need to know the radius \( r \) or the speed \( v \) to compute the exact values. If you have a specific radius or speed, please provide it, and I can help you calculate the exact values for \( v_i \), \( v_f \), and the distances OP for both ions.