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 are 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 desires 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 400 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

a) Magnetic force on the particles is given by,

F = q(v x B)

Since velocity is to the right just as the particle enters the field and magnetic field B is directed out of the page, the direction of (v x B) is downwards. Since the paths of the particles are curved down, the force must be acting downward at the entrance point. Hence, we can say that the particles are positively charged.

Since these positive charged particles are moving towards the right plate, we can say that the right plate is at a lower potential. So, the right plate is at higher potential.

b) Potential energy of the particles is converted to their Kinetic energy when the particles accelerates through the potential difference. So,

q(V) = mv2/2

=> v = (2qV/m)1/2

Radius of the orbit, r = mv/Bq = m * (2qV/m)1/2 / Bq

=> r = (2mV/q)1/2/B

So, d = 2r = (8mV/q)1/2/B

For proton,

d = [8 * (1.67 * 10-27) * 400 / (1.6 * 10-19)]1/2 / (!20 * 10-3) = 4.8 cm

c) For deuterium nucleus,

d = [8 * (2 * 1.67 * 10-27) * 400 / (1.6 * 10-19)]1/2 / (!20 * 10-3) = 6.8 cm

So, the detector needs to be moved down by (6.8 - 4.8) cm = 2 cm

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