A beam of charged particles consisting of protons, electrons, deuterons, singly-

Question

A beam of charged particles consisting of protons, electrons, deuterons, singly-ionized helium atoms, and singly ionized H2 molecules all pass through what is known as a velocity selector. Singly ionized means that the atom or molecule has lost one electrons so its charge is -1e.
All of the particles emerge with the same velocity of 2.5E6 m/s. The beam of charged particles then enters a region of uniform magnetic field where B=0.40 Tesla directed perpendicular to their velocity.
a) Compute the radius of curvature for the path of each type of particle.
b) Explain briefly why this information could be important?
Hint: Re-consider Thompson’s 1st measurement of e/m and what is different among each particle. Also, 1 amu = 1.66E-27 kg or 931 MeV/c^2

Explanation / Answer

The force due to the magnetic field is equal to the force acting on the particle due to its circular motion (centripetal force)      F = B q V      F = mV^2 /R Equating the equations    B q V = MV^2 / R The radius of path of the particle is     R = mV / B q Here q = 1e = 1.6x10^-19 C, B = 0. 4 T, V = 2.5x1-^6m/s    R = m (2.5x10^6) / (0.4)(1.6x10^-19)       = m (3.9x10^25 ) The radius of hte path depends upon the mass of hte particles. For electrom m = 9.1x 10^-31 kg For proton m = 1.67x10^-27 kg ........................ On substituting the values for m, we can get the radius of path followed by the particles. ------------------------------------------------------------------------------------------------- b) The radius of the circular path can be defined as       R = V / B [ e/m ] As V and B are constant, the radius of the path depends upon the e/m of the particle.    e/ m is called specific charge. It is constant for each particle. So the path radius depends upon the specific charge of the particle. Hope this is clear.

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