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(A) The frequency of revolution $
u$ of a charged particle moving in a magnetic field is given by the formula: $
u = \frac{qB}{2\pi m}$.Since the ma

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$Li^+$

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(A) The frequency of revolution $
u$ of a charged particle moving in a magnetic field is given by the formula: $
u = \frac{qB}{2\pi m}$.Since the magnetic field $B$ is constant and the velocity is the same,the frequency $
u$ is directly proportional to the charge-to-mass ratio: $
u \propto \frac{q}{m}$.To find the minimum frequency,we need to find the particle with the minimum $\frac{q}{m}$ ratio.For $Li^+$: $q = +1e$,$m \approx 7u$,so $\frac{q}{m} \approx \frac{1}{7}$.For Electron: $q = -1e$,$m \approx \frac{1}{1836}u$,so $\frac{q}{m} \approx 1836$.For Proton: $q = +1e$,$m \approx 1u$,so $\frac{q}{m} = 1$.For $He^+$: $q = +1e$,$m \approx 4u$,so $\frac{q}{m} = 0.25$.Comparing the ratios,$Li^+$ has the smallest $\frac{q}{m}$ value. Therefore,$Li^+$ will have the minimum frequency of revolution.

Which particles will have the minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?

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