Question

significant figures, what are the cyclotron frequencies in a magnetic field of the ions (a)${\mathrm{O}}_2^{+}$, (b)${\mathrm{N}}_2^{+}$and (c)${\mathrm{CO}}^{+}$? The atomic masses are shown in the table; the mass of the missing electron is less than $0.001u$and is not relevant at this level of precision. Although ${\mathrm{N}}_7^{+}\text{and}{\mathrm{CO}}^{+}$both have a nominal molecular mass of $28$, they are easily distinguished by virtue of their slightly different cyclotron frequencies. Use the following constants:$1\mathbf{u}=1.6605\times {10}^{-27}\mathrm{kg},e^e=1.6022\times {10}^{-19}\mathrm{C}.$

Short answer

(a) The magnetic field of the ions$f_{{\mathrm{O}}_2^{+}}=1.4401\mathrm{MHz}$

(b) The magnetic field of the ions $f_{{\mathrm{N}}_2^{+}}=1.6450\mathrm{MHz}$

(c) The magnetic field of the ions $f_{{\mathrm{CO}}^{+}}=1.6457\mathrm{MHz}$

Step-by-step solution

Introduction

The electromagnetic influence on electric currents, pulsed electrical, and magnetic materials is characterized by a magnetic field, that is a vector field. In a magnetic field, a strong growth experiences a force that's also perpendicular to both its own motion and the magnetic field.

Find the magnetic field

(a) The cyclotron mobility is the circular motion of a particle subjected to a magnetic field at a constant speed. The magnetic field leads the circular motion to generate a circular frequency, which is given by the following ($29.21$) in the format

Equation $1$

$f_{\mathrm{cyc}}=\frac{qB}{2\pi m}$

While $q$represents a photon's charge, $B$the magnetic dipole, and $m$the particle's mass. The frequency is determined by the magnetism rather than the particle's velocity, as shown in equation ($1$).

$m=2\left(15.995\mathrm{u}\right)=31.99\mathrm{u}$is the chemical mass of the anion. Let's use $kg$to calculate the mass.

$m_{{\mathrm{O}}_2^{+}}=\left(31.99\mathrm{u}\right)\left(\frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right)=5.3119\times {10}^{-26}\mathrm{kg}$

So we fill in the quantities for the electron's$B,m_{{\mathrm{O}}_2^{+}}\text{and}q$can get $f_{{\mathrm{O}}_2^{+}}$

$\begin{array}{r}{f}_{{\mathrm{O}}_2^{+}}=\frac{q{B}_{{\mathrm{O}}_2^{+}}}{2\pi m_{{\mathrm{O}}_2^{+}}}\\ =\frac{\left(1.6022\times {10}^{-19}\mathrm{C}\right)\left(3.0000\mathrm{T}\right)}{2\pi \left(5.3119\times {10}^{-26}\mathrm{kg}\right)}\\ =1.4401\times {10}^6\mathrm{Hz}\\ =1.4401\mathrm{MHz}\end{array}$

Find the magnetic field

(b) The nitrogen ion has a molecular mass of $m=2\left(14.003\mathrm{u}\right)=28.006\mathrm{u}$. Let's convert the mass to kilogrammes.

$m_{{\mathrm{N}}_2^{+}}=\left(28.006\mathrm{u}\right)\left(\frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right)=4.6504\times {10}^{-26}\mathrm{kg}$

Now we fill in the values for the electron's $B,m_{{\mathrm{N}}_2^{+}}\text{and}q$to get $f_{{\mathrm{N}}_2^{+}}$

$\begin{array}{r}{f}_{{\mathrm{N}}_2^{+}}=\frac{q{B}_{{\mathrm{N}}_2^{+}}}{2\pi m_{{\mathrm{N}}_2^{+}}}\\ =\frac{\left(1.6022\times {10}^{-19}\mathrm{C}\right)\left(3.0000\mathrm{T}\right)}{2\pi \left(4.6504\times {10}^{-26}\mathrm{kg}\right)}\\ =1.6450\times {10}^6\mathrm{Hz}\\ =1.6450\mathrm{MHz}\end{array}$

Find the magnetic field

(c) The carbon monoxide ion's molar weights is $\left(12.000\mathrm{u}\right)+\left(15.995\mathrm{u}\right)=27.995\mathrm{u}$. Let us just convert the value to$kg$using the formula below.

$m_{{\mathrm{CO}}^{+}}=\left(27.995\mathrm{u}\right)\left(\frac{1.6605\times {10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right)=4.6485\times {10}^{-26}\mathrm{kg}$

Finally we enter in the values for the electron's $B,m_{{\mathrm{CO}}^{+}}\text{and}q$to get$f_{{\mathrm{CO}}^{+}}$

$\begin{array}{r}{f}_{{\mathrm{CO}}^{+}}=\frac{q{B}_{{\mathrm{CO}}^{+}}}{2\pi m_{{\mathrm{CO}}^{+}}}\\ =\frac{\left(1.6022\times {10}^{-19}\mathrm{C}\right)\left(3.0000\mathrm{T}\right)}{2\pi \left(4.6485\times {10}^{-26}\mathrm{kg}\right)}\\ =1.6457\times {10}^6\mathrm{Hz}\\ =1.6457\mathrm{MHz}\end{array}$