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.
With what speed u does the ion exit the acceleration region?
Find the speed in terms of m, q, V, and any constants.

u =

Answer 1

In a mass spectrometer, the speed at which an ion exits the acceleration region is determined by the equation u = √(2qV/m), where u is the speed, q is the charge, V is the potential difference, and m is the mass of the ion.

Step-by-step explanation:

J. J. Thomson’s significant contributions to physics include the discovery of the electron and the development of the mass spectrometer. A key application of the mass spectrometer is to determine the ratio of mass m to charge q of an ion.

The basic operation of a mass spectrometer involves accelerating an ion through a potential difference V, resulting in the ion gaining kinetic energy.

The kinetic energy acquired by the ion is equal to the charge times the potential difference, expressed as qV.

When using the principles of conservation of energy, the final kinetic energy of the ion when it exits the acceleration region can be calculated using the equation K.E. = (1/2)mu² = qV, where u is the speed of the ion, m is the mass of the ion, and V is the potential difference.

The speed u of the ion is then given by:

u = √(2qV/m)