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A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses. In one design for such an instrument, ions with mass m and charge are accelerated through a potential difference V. They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius . A detector measures where the ions complete the semicircle and from this it is easy to calculate . (a) Derive the equation for calculating the mass of the ion from measurements of , and. (b) What potential difference V is needed so that singly ionized 12 atoms will have R = 50.0 cm in a 0.150T magnetic field? (c) Suppose the beam consists of a mixture of and ions. If v and B have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished?

Step by step solution

01

Definition of isotopes

The term isotopes may be defined as the element having same atomic number and different mass number.

02

Determine equation for calculating the mass of the ion from measurements of B, V , R and q . 

For calculate velocity we equate kinetic energy and potential energy

$\begin{array}{r}qV=\frac{1}{2}m{v}^2\\ v=\sqrt{\frac{2qV}{m}}\end{array}$

And the radius of curvature

$$\begin{gathered} \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}} \\ \mathrm{Using}\mathrm{above}\mathrm{velocity}\mathrm{result} \\ \begin{array}{r}\mathrm{R}=\frac{\mathrm{m}\sqrt{2\mathrm{qV}/\mathrm{m}}}{\mathrm{qB}}\\ \mathrm{m}=\frac{{\mathrm{qB}}^2{\mathrm{R}}^2}{2\mathrm{V}}\end{array} \end{gathered}$$

Hence, the equation for calculating the mass of the ion from measurements of , and q is $m=\frac{q{B}^2{R}^2}{2V}$

From the above result

$$\begin{gathered} \mathrm{V}=\frac{{\mathrm{qB}}^2{\mathrm{R}}^2}{2\mathrm{m}} \\ \mathrm{Put}\mathrm{given}\mathrm{values} \\ \begin{array}{r}\mathrm{V}=\frac{\left(1.60脳{10}^{鈭�19}\mathrm{C}\right)(0.150\mathrm{T}{)}^2(0.500\mathrm{m}{)}^2}{2\left(12\right)\left(1.66脳{10}^{鈭�27}\mathrm{kg}\right)}\\ \mathrm{V}=2.26脳{10}^4\mathrm{V}\end{array} \end{gathered}$$

Hence, the potential difference V is needed so that singly ionized 12 C atoms will have R = 50.0 cm in a 0.150 T magnetic field is$V=2.26脳{10}^{4}V$.

The separation of two isotopes can be calculate as

$\begin{array}{r}\mathrm{螖}D=D_{14}鈭�D_{12}\\ \mathrm{螖}D=2R\\ \mathrm{螖}D=2\sqrt{\frac{2Vm}{q{B}^2}}\end{array}$

So

$\begin{array}{r}\mathrm{螖顿}=2\sqrt{\frac{2\mathrm{Vm}}{{\mathrm{qB}}^2}}鈭�2\sqrt{\frac{2\mathrm{Vm}}{{\mathrm{qB}}^2}}\\ \mathrm{螖顿}=2\sqrt{\frac{2\mathrm{V}\left(1\mathrm{u}\right)}{\mathrm{qB}}(\sqrt{14}鈭�\sqrt{12})}\\ \mathrm{螖顿}=2\sqrt{\frac{2\left(2.26脳{10}^4\mathrm{V}\right)\left(1.66脳{10}^{鈭�27}\mathrm{kg}\right)}{\left(1.66脳{10}^{鈭�19}\mathrm{C}\right)(0.150\mathrm{T}{)}^2}(\sqrt{14}鈭�\sqrt{12})}\\ \mathrm{螖顿}=8.01脳{10}^{鈭�2}\mathrm{m}\end{array}$

Hence, the separation of these two isotopes is $8.01脳{10}^{鈭�2}\mathrm{m}$.

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