Two positive ions, labeled A and B, are accelerated from rest through the same e

Question

Two positive ions, labeled A and B, are accelerated from rest through the same electric potential difference delta V. In the process, particle A acquires a speed VA and particle B acquires a speed VB. The mass of the particle A is a denoted MA and the mass of particle B is denoted MB, while the charge of the particle A is denoted qA and the mass of particle B is denoted qB.

a) Derive a symbolic expression for the ratio of velocities, VB/VA, of the ions in terms of the variables given

b) Derive a symbolic expression for that ratio of kinetic energies, KEB/KEA, of the ions in terms of the variables given. The ions are moving sufficiently slowly so the kinetic energy for either ion can be expressed as KE=1/2mv^2

c) Derive a symbolic expression for the ratio of potential energies, PEB/PEA, of the ions in terms of the variables given.

d) Derive a symbolic expression for the ratio of total energies. EB/EA, of the ions in terms of the variables give. Recall the total energy= KE+PE.

e) evaluate your answer for part (d) above assuming that ion "A" is C+ (carbon ion) so that MA=6u and QA= +e (where u is an atomic mass unit and e is a fundamental charge) and assuming that ion "B" is O2+ (oxygen ion, doubly charged) so that MB=8u and QB=+2e

Explanation / Answer

kea/keb = [ma*va^2]/[mb*vb^2] pea/peb = [ma*g*xa]/[mb*g*xb] = [ma*xa]/[mb*xb] ea / eb = ([0.5*ma*va^2] + [ma*g*xa])/([0.5*mb*vb^2] + [mb*g*xb])

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