Spectrometers are useful analysis tools that are used across many branches of sc

Question

Spectrometers are useful analysis tools that are used across many branches of science (and you might have heard the term on CSI or other crime shows). Some spectrometers arc mass spectrometers and separate different atoms based on their q/m ratios. Others are able to separate otherwise identical charged particles, like electrons, that have different velocities. A simple spectrometer model is shown below. Charged particles are first accelerated from rest through an electric field and then injected into an area with a uniform magnetic field. Based on their masses, charges, or velocities, the particles will experience different paths in the magnetic field. A detector can then be placed at the desired location to collect the specific particles you are interested in. a. Based on the scenario above, which plate is at a higher potential, the left plate or the right plate? Justify your answer. b. The magnetic field in a spectrometer given by the diagram above has a magnitude of 120 mT. If the potential difference between the plates is 4(X) V, how far from where a proton exits the plates would you need to place a detector to detect protons? c. As you may know, a single proton is also a hydrogen nucleus. Deuterium is like hydrogen but the nucleus is replaced by a proton and a neutron. If you want to detect a deuterium nucleus, how far would you need to move the detector from its position in Part (b) and in what direction?

Explanation / Answer

Part a

As magnetic field is perpendicular to the plane of paper directed outwards,

It means from right hand curl rule, a positive particle will turn downward and a negative particle will turn upward in figure.

As given particle turns downward in figure, So given particle has positive charge.

Now, as particle has positive charge and accelerated towards right by the plate, it means left plate is at higher potential than right plate.

Part b

Magnetic field, B=120 mT =120 * 10-3T

Potential difference, T=400V

Mass of proton, m=1.67*10-27 Kg

Charge on proton, q=1.6*10-19 C

Now,

            As     mv2 /r = Bqv   

And d=2r

So                    2mv2/d =Bvq

                        d=2mv/Bq      …………………………………(1)

Protons kinetic energy when it entered into magnetic field is

                        mv2/2 =q(V)

so                    v=sqrroot(2qV/m)   ……………………………………..(2)

v = sqrroot(2*1.6*10-19 *400 /1.67*10-27)=2.7685 * 105 m/s

so from ……………(1)

                        d= 2mv/Bq =2*1.67*10-27 * (2.7685*105) /(120*10-3 *1.6*10-19)

                        d=0.04614 m = 4.614cm

so required distance is d=4.614 cm.

Part c

Mass of deuterium nucleus, m=3.35*10-27 kg

So from ………………………..(2)

                        v=sqrroot(2qV/m)  

                        v = sqrroot(2*1.6*10-19 *400 /3.35*10-27)=1.955*105 m/s

now from …………………….(1)

                        d= 2mv/Bq =2*3.35*10-27 * (1.955*105) /(120*10-3 *1.6*10-19)

                        d=0.06822m =6.822cm.

so required distance is d=6.822 cm. and detector should be placed downward as shown in given figure.

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