A mass spectrometer (see diagram) is an analytic device used to identify various

Question

A mass spectrometer (see diagram) is an analytic device used to identify various charged particles by measuring their charge to mass ratio. The charged particles are accelerated through a potential difference Delta V, 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 the accelerating voltage is slowly increased or decreased, different charged particles reach the detector and are measured. The diagram shows the set-up for detecting positive charges, but it can be easily adjusted to detect negative charges (by reversing the polarity of the accelerating potential and reversing the direction of the B-field) So we'll assume that the necessary adjustments have been made in this question if we are talking about negative charges rather than positive ones. Assume the B-field is at B_0 = 8 times 10^-4 T and the spacing between the entrance and exit holes is d = 8.00 cm. If the accelerating potential is adjusted to 500V, with what speed (in d m/s) do the electrons get ejected into the magnetic field? a. 1.33 times 10^7 b. 2.94 times 10^7 c. 0.81 times 10^7 d. 7.11 times 10^7 How far away (in cm) from the entrance hole would those electrons strike the wall? a. 3.55 b. 6.31 c. 18.9 d. 23.7 What magnetic field (in mu T) would have to be applied to allow the electrons to be detected by the detector? a. 511 b. 2225 c. 1886 d. 800 If the magnetic field is returned back to its original magnitude, B_0, what accelerating potential (in V) is necessary to detect protons? a.6.77 b. 722 c. 85.3 d. 0.0491 If the magnetic field remains at its original magnitude, B_0, but the potential is adjusted to 200V, what will be the charge-to-mass ratio (in C/kg) of the detected particles? a. 8.99 times 10^11 b. 3.91 times 10^11 c. 1.33 times 10^11 d. 13.6 times 10^11

Explanation / Answer

27) from the formula eV= mv2 /2

here V is potential and v is velocity

so v=sqrt( 2*e*V/m)----------------------------1)

here e=1.6*10-19 , V=500v and m=9.1*10-31 kg

putting all these values in equation 1) we get

v=1.33*107 m/sec answer

29 )from the formla B=mv/er ,putting the values

m=9.1*10-31 kg r=4cm=4*10-2 m

putting the values

we get B=1886 micro T Answer

31) e/m= 2V/(B2 r2 )

putting the value of V=500V, B=8*10-4 T, r=4*10-2 m

putting the values

we get e/m= 8.99*1011

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