Question

Mass Spectrometer J. J. Thomson is best known for his discoveries about the nature of cathode rays. Another important contribution of his was the invention, together with one of his students, of the mass spectrometer. The ratio of mass m to (positive) charge q of an ion may be accurately determined in a mass spectrometer. In essence, the spectrometer consists of two regions: one that accelerates the ion through a potential difference V and a second that measures its radius of curvature in a perpendicular magnetic field. (Figure 1) The ion begins at potential V and is accelerated toward zero potential. When the particle exits the region with the electric field it will have obtained a speed u.

Part A With what speed u does the ion exit the acceleration region? Find the speed in terms of m, q, V, and any constants. u = Part B After being accelerated, the particle enters a uniform magnetic field of strength B0 and travels in a circle of radius R (determined by observing where it hits on a screen--as shown in the figure). The results of this experiment allow one to findm/q in terms of the experimentally measured quantities such as the particle radius, the magnetic field, and the applied voltage. What is m/q? Express m/q in terms of B0,V, R, and any necessary constants. m/q =

Answer 1

Its given that,

mass of the particle \(=\mathrm{m}\), Charge \(=\mathrm{q}\) and Potential \(=\mathrm{V}\)

Mag field \(=\mathrm{B} 0\) and its travelling in the circular path of radius \(=\mathrm{R}\)

Part (A)With what speed u does the ion exit the acceleration region?

Let E1 be the intial energy of the particle which is actually electirc potential energy and given by

\(E 1=q V\)

When the particle is moving its energy gets converted into \(\mathrm{KE}\) and eguals to

\(\mathrm{E} 2=1 / 2 \mathrm{~m} \mathrm{u}^{2}\)

Using conservation of energy, we get

\(q V=1 / 2 m u^{2}\) this will give us

\(\mathrm{u}=\sqrt{(2 q V) / m}\)

\(\operatorname{Part}(B)\)

The cyclotron frequency, which is the (angular) frequency of the orbital motion of the ion in the magnetic field. Its given by

\(\omega=q \mathrm{BO} / \mathrm{m}\)

By the defination og angular speed we know that,

\(\mathrm{u}=\mathrm{R} \omega \mathrm{So} \omega=\mathrm{u} / \mathrm{R}\)

Equating the two diff expressions of \(\omega\), we get

q \(\mathrm{BO} / \mathrm{m}=\mathrm{u} / \mathrm{R}\)

Putting the value of \(\mathrm{u}\), we have obtained in Part (a)

q \(\mathrm{BO} / \mathrm{m}=\sqrt{(2 q V) / m} / \mathrm{R}\)

Squaring both th sides we get,

\(q^{2} B 0^{2} / m^{2}=(2 q V / m) / R^{2}\)

Hence we get, \(m / q=R^{2} B 0^{2} /(2 V)\)

Answer 2

a)According to the law of conservation of energy we haveqV = (1/2)mv²b)Now when the the ion enters into the magnetic field then itfollows the circular path of radius(R), its formula is