91影视

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.

(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 $1$

$f_{\mathrm{cyc}}=\frac{qB}{2蟺m}$

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.

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

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

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

(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.

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

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

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

(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.

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

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

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