Q. 64 FIGURE聽P29.64聽shows a mass spe... [FREE SOLUTION]

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Chapter 29: Q. 64 (page 834)

$F I G U R E P 29.64$ shows a mass spectrometer, an analytical instrument used to identify the various molecules in a sample by measuring their charge-to-mass ratio $\frac{q}{m}$ . The sample is ionized, the positive ions are accelerated (starting from rest) through a potential differencelocalid="1648976601527" $鈭� V$ , 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 localid="1648976606181" $200.00 m T$ a magnetic field and an localid="1648976610307" $8.0000 c m$ spacing between the entrance and exit holes. To five significant figures, what accelerating potential differences localid="1648978768434" $鈭� V$ are required to detect the ions localid="1648978902862" $(a) O_{2}^{+}, (b) N_{2}^{+},$ and localid="1648978898077" $(c) {CO}^{+}$ ? See Exercise localid="1648978753549" $29$ for atomic masses; the mass of the missing electron is less than localid="1648978758219" $0.001 u$ and is not relevant at this level of precision. Although localid="1648978910549" $N_{2}^{+}$ and localid="1648978778238" ${CO}^{+}$ both have a nominal molecular mass of localid="1648978782098" $28$ , they are easily distinguished by virtue of their slightly different accelerating voltages. Use the following constants:
$1 u = 1.6605 脳 10^{鈭� 27} kg, e = 1.6022 脳 10^{鈭� 19} C .$

Short Answer

Expert verified

Part $(a)$

$(a)$ For $O_{2}^{+}$ ,the potential difference is $螖 V = 96.519 V .$

Part $(b)$

$(b)$ For $N_{2}^{+}$ ,the potential difference is $螖 V = 110.25 V .$

Part $(c)$

$(c)$ For ${CO}^{+}$ ,the potential difference is $螖 V = 110.29 V .$

Step by step solution

Step: 1 Finding potential difference:

As each ion is lacking one electron, it has a positive charge of magnitude $e$ . At the entry hole to the area with uniform magnetic field, the work done by the electric force $e 鈭� V$ is transformed to the kinetic energy of the ion:

$\frac{1}{2} m v^{2} = e 螖 V v = \sqrt{\frac{2 e 螖 V}{m}}$

Because the magnetic force is always perpendicular to the direction of motion, the ion's speed remains constant after it reaches the uniform magnetic field. It works as a centripetal force, bending the ion's course to be round.

$\frac{m v^{2}}{r} = e v B$

where localid="1648978843687" $d = 2 r$ (localid="1648978848816" $r$ is the trajectory's radius and d is the diameter) such that we can write

localid="1648978203871" $\frac{2 m v^{2}}{d} = e v B 鈬� \frac{2 m v}{d} = e B \frac{2 m}{d} \sqrt{\frac{2 e 螖 V}{m}} = e B \frac{4 m^{2}}{d^{2}} \frac{2 e 螖 V}{m} = e^{2} B^{2} 鈬� \frac{8 m}{d^{2}} 螖 V = e B^{2} 螖 V = \frac{e B^{2} d^{2}}{8 m} .$

Step: 2 Equating the equations:

Each ion's mass in relation of atomic weight is

$A (O) = 15.995, A (N) = 14.003, A (C) = 12.000 \begin{array}{l} m (O_{2}^{+}) = 2 A (O) u \\ m (N_{2}^{+}) = 2 A (N) u \\ m ({CO}^{+}) = (A (C) + A (O)) u . \end{array}$