Please answer the question, question is in the image attached. please answer all the parts 27, 28, 29, 30 and 31. Thanks! A mass spectrometer (see diagram) is an analytic device used to identify various charged particles bymeasuring their charge to mass ratio. The charged particles are accelerated through a potential differenceAV, and they enter a region of uniform magnetic field. The field bends the ions into circular trajectories,but after just half a circle they either strike the wall or pass through a small opening to a detector. As theaccelerating voltage is slowly increased or decreased, different charged particles reach the detector and aremeasured. The diagram shows the set-up for detecting positive charges, but it can be easily adjusted todetect negative charges (by reversing the polarity of the accelerating potential and reversing the direction ofthe B-field). So we'll assume that the necessary adjustments have been made in this question if we aretalking about negative charges rather than positive ones.AVAssume the B-field is at Bo = 8x104 T and the spacing between the entranceand exit holes is d=8.00 cm.27. If the accelerating potential is adjusted to 500V, with what speed (inm/s) do the electrons get ejected into the magnetic field? Ans: Aa. 1.33x107b. 2.94x107c. 0.81x10d. 7.11x10"Detector28. How far away (in cm) from the entrance hole would those electronsstrike the wall? Ans: Ca. 3.55b. 6.31с. 18.9d. 23.729. What magnetic field (in µT) would have to be applied to allow the electrons to be detectedby the detector? Ans: Ca.511b. 2225c. 1886d. 80030. If the magnetic field is returned back to its original magnitude, Bo, what acceleratingpotential (in V) is necessary to detect protons? Ans: Db. 722a. 6.77c. 85.3d. 0.049131. If the magnetic field remains at its original magnitude, Bo, but the potential is adjusted to200V, what will be the charge-to-mass ratio (in C/kg) of the detected particles? Ans: Ba. 8.99x10"b. 3.91x10"c. 1.33x10"е. 13.6х101

We’ll answer the first question since the exact one wasn’t specified. Please submit a new question specifying the one you’d like answered.It is given that: Bo=8×10-4 Td=8.00 cm27)The change in kinetic energy is q∆V12mV2-0=q∆VV=2q∆VmV=2×1.6×10-19×5009.1×10-31V=1.33×107 msHence, the correct option is V=1.33×107 ms