$FIGUREP29.64$shows a mass spectrometer, an analytical instrument used to identify the various molecules in a sample by measuring their charge-to-mass ratio $\raisebox{1ex}{$q$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$. The sample is ionized, the positive ions are accelerated (starting from rest) through a potential differencelocalid="1648976601527" $∆V$, 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" $200.00mT$a magnetic field and an localid="1648976610307" $8.0000cm$spacing between the entrance and exit holes. To five significant figures, what accelerating potential differences localid="1648978768434" $∆V$are required to detect the ions localid="1648978902862" $\text{(a)}{\mathrm{O}}_2^{+}\text{, (b)}{\mathrm{N}}_2^{+},$ and localid="1648978898077" $\left(\mathrm{c}\right){\mathrm{CO}}^{+}$? See Exercise localid="1648978753549" $29$for atomic masses; the mass of the missing electron is less than localid="1648978758219" $0.001u$and is not relevant at this level of precision. Although localid="1648978910549" ${\mathrm{N}}_2^{+}$and localid="1648978778238" ${\mathrm{CO}}^{+}$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:

$1\mathrm{u}=1.6605\times {10}^{-27}\mathrm{kg},e=1.6022\times {10}^{-19}\mathrm{C}.$

Step by step solution

01

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 $e∆V$is transformed to the kinetic energy of the ion:

$$\begin{gathered} \frac{1}{2}m{v}^2=e\mathrm{\Delta }V \\ v=\sqrt{\frac{2e\mathrm{\Delta }V}{m}} \end{gathered}$$

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.

$\frac{m{v}^2}{r}=evB$

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

localid="1648978203871" $$\begin{gathered} \frac{2m{v}^2}{d}=evB\Rightarrow \frac{2mv}{d}=eB \\ \frac{2m}{d}\sqrt{\frac{2e\mathrm{\Delta }V}{m}}=eB \\ \frac{4m^2}{d^2}\frac{2e\mathrm{\Delta }V}{m}=e^2{B}^2 \\ \Rightarrow \frac{8m}{d^2}\mathrm{\Delta }V=e{B}^2 \\ \mathrm{\Delta }V=\frac{e{B}^2{d}^2}{8m}. \end{gathered}$$

02

Step: 2 Equating the equations:

Each ion's mass in relation of atomic weight is

$$\begin{gathered} A\left(\mathrm{O}\right)=15.995,A\left(\mathrm{N}\right)=14.003,A\left(\mathrm{C}\right)=12.000 \\ \begin{array}{l}m\left({\mathrm{O}}_2^{+}\right)=2A\left(\mathrm{O}\right)u\\ m\left({\mathrm{N}}_2^{+}\right)=2A\left(\mathrm{N}\right)u\\ m\left({\mathrm{CO}}^{+}\right)=\left(A\right(\mathrm{C})+A(\mathrm{O}\left)\right)u.\end{array} \end{gathered}$$

03

Step: 3 Obtaining the values: (part a and part b and part c)

Getting potential difference by substituting the values in equation,

For ${\mathrm{O}}_2^{+}$,

role="math" localid="1648978509176" $\mathrm{\Delta }V=\frac{e{B}^2{d}^2}{16A\left(\mathrm{O}\right)u}=96.519\mathrm{V}.$

For ${\mathrm{N}}_2^{+}$,

$\mathrm{\Delta }V=\frac{e{B}^2{d}^2}{16A\left(\mathrm{N}\right)u}=110.25\mathrm{V}.$

For ${\mathrm{CO}}^{+}$,

$\mathrm{\Delta }V=\frac{e{B}^2{d}^2}{8\left(A\right(\mathrm{C})+A(\mathrm{O}\left)\right)u}=110.29\mathrm{V}.$

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