(Figure 1) shows a mass spectrometer, an analytical instrument used to identify
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Question
(Figure 1) shows a mass spectrometer, an analytical instrument used to identify the various molecules in a sample by measuring their charge-to-mass ratio q/m. The sample is ionized, the positive ions are accelerated (starting from rest) through a potential difference V, and they then 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, different ions reach the detector and are measured. Consider a mass spectrometer with a 200.00 mT magnetic field and an 8.0000 cm spacing between the entrance and exit holes.
To five significant figures, what accelerating potential difference V is required to detect the ion O+2? The masses of the atoms are shown in the table; the mass of the missing electron is less than 0.001 u and is not relevant at this level of precision. Use the following constants: 1 u = 1.6605×1027kg, e = 1.6022×1019C.
Atomic masses
Express your answer to five significant figures and include the appropriate units.
what is Vo2
What accelerating potential difference V is required to detect N+2?
Express your answer to five significant figures and include the appropriate units.
Although N+2 and CO+ both have a nominal molecular mass of 28, they are easily distinguished by virtue of their slightly different accelerating voltages. What accelerating potential difference V is required to detect CO+?
Express your answer to five significant figures and include the appropriate units.
Vco+
12C 12.000 u 14N 14.003 u 16O 15.996 u AV DetectorExplanation / Answer
In magnetic field ,Bvq = mv 2 / r
Bq = mv / r
From this speed of the ion v = Bqr / m
Where q = charge of ion = 2e = 2x 1.6022×1019C
m = mass of ion = 15.966 u =15.966 x1.6605×1027kg
B = magnetic field = 200 mT = 0.2 T
r = radius = separation of enterance and exit / 2
= 8 cm / 2 = 4 cm = 0.04 m
Substitute values you get ,
v = (0.2)(2x 1.6022×1019)(0.04) /(15.966 x1.6605×1027)
= 96694.48 m/s
We know (1/2)mv 2 = Vq
From this potential difference V = (1/2q)mv 2
V = (15.966 x1.6605×1027)(96694.48) 2 / [2x(2x 1.6022×1019)]
= 386.78 volt
Similarly you do remaining
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