Question

You are using a type of mass spectrometer to measure charge-to-mass ratios of atomic ions. In the device, atoms are ionized with a beam of electrons to produce positive ions, which are then accelerated through a potential difference V. (The final speed of the ions is great enough that you can ignore their initial speed.) The ions then enter a region in which a uniform magnetic field is perpendicular to the velocity of the ions and has magnitude B = 0.250 T. In this region, the ions move in a semicircular path of radius . You measure as a function of the accelerating voltage V for one particular atomic ion:

(a) How can you plot the data points so that they will fall close to a straight line? Explain. (b) Construct the graph described in part (a). Use the slope of the best-fit straight line to calculate the charge-to-mass ratio for the (q/m) ion. (c) For V = 12.0kV , what is the speed of the ions as they enter the region? (d) If ions that have R = 21.2cm for V = 12.0 kV are singly ionized, what is when V = 12.0 kV for ions that are doubly ionized?

Short answer

The relation between R and$\sqrt{\mathrm{V}}\text{is}\mathrm{R}=\left(\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{m}}{\mathrm{q}}}\right)\sqrt{\mathrm{V}}$ is straight line with slope $(1/\mathrm{B})\sqrt{2\mathrm{m}/\mathrm{q}}$.

The average charge to mass ratio is$\frac{\mathrm{q}}{\mathrm{m}}=8.0\times {10}^6\mathrm{C}/\mathrm{kg}$ .the speed of ions is$5.63\times {10}^5\mathrm{m}/\mathrm{s}$ and the value of R is${\mathrm{R}}_{2\mathrm{q}}=15.0\mathrm{cm}$ .

Step-by-step solution

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

Determine the straight-line equation, construct graph, speed of ions and value of R

$\mathrm{R}=\left(0.002\mathrm{m}/{\mathrm{V}}^{1/2}\right)\sqrt{V}-\left(0.0028\mathrm{cm}\right)$

When ions enter in magnetic field the radius of the circular path can be obtain from the result localid="1668250514496" $\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}$and from the kinetic energy and potential energy we write localid="1668250517134" $\mathrm{v}=\sqrt{\frac{2\mathrm{qV}}{\mathrm{m}}}$,so

localid="1668250519459" $\mathrm{R}=\frac{1}{\mathrm{B}}\left(\sqrt{\frac{2\mathrm{m}}{\mathrm{q}}}\right)\sqrt{\mathrm{V}}$.

Which is the relation between R and localid="1668250522234" $\mathrm{R}=\left(\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{m}}{\mathrm{q}}}\right)\sqrt{\mathrm{V}}$is straight line with slope localid="1668250524574" $(1/\mathrm{B})\sqrt{2\mathrm{m}/\mathrm{q}}$.

Now the graph plot as

The best fitting equation is and the sloe is $\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{m}}{\mathrm{q}}}=0.002\mathrm{m}/{\mathrm{V}}^{1/2}$

Or

$\begin{array}{r}\frac{\mathrm{q}}{\mathrm{m}}=\frac{2}{\left[\mathrm{B}\right(\text{slope}){]}^2}\\ =\frac{2}{{\left[(0.250\mathrm{T})\left(0.002\mathrm{m}\cdot {\mathrm{V}}^{-1/2}\right)\right]}^2}\\ =8.0\times {10}^6\mathrm{C}/\mathrm{kg}\end{array}$

Hence, the average charge to mass ratio is $\frac{\mathrm{q}}{\mathrm{m}}=8.0\times {10}^6\mathrm{C}/\mathrm{kg}$.

The speed can be calculated from the relation

$\begin{array}{r}\mathrm{v}=\sqrt{\frac{2\mathrm{qV}}{\mathrm{m}}}\\ =\sqrt{2\left(20.0\times {10}^3\mathrm{V}\right)\left(7.924\times {10}^6\mathrm{C}/\mathrm{kg}\right)}\\ =5.63\times {10}^6\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of ions is $5.63\times {10}^5\mathrm{m}/\mathrm{s}$.

The value of ${\mathrm{R}}_{2\mathrm{q}}$obtain by the relation

$\begin{array}{r}{\mathrm{R}}_{\mathrm{q}}=\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{mV}}{\mathrm{q}}}\\ {\mathrm{R}}_{2\mathrm{q}}=\frac{1}{\mathrm{B}}\sqrt{\frac{2\mathrm{mV}}{2\mathrm{q}}}\\ {\mathrm{R}}_{2\mathrm{q}}=\frac{{\mathrm{R}}_{\mathrm{q}}}{\sqrt{2}}\\ \mathrm{Then}\mathrm{put}\mathrm{values}\\ {\mathrm{R}}_{2\mathrm{q}}=\frac{21.2\mathrm{cm}}{\sqrt{2}}\\ {\mathrm{R}}_{2\mathrm{q}}=15.0\mathrm{cm}\end{array}$

Hence, the value of R is${\mathrm{R}}_{2\mathrm{q}}=15.0\mathrm{cm}$.