Which of the following particles will exhibit the minimum frequency...

The frequency of revolution of a charged particle in a uniform magnetic field is given by $f = \dfrac{qB}{2\pi m}$.

This shows that the frequency is independent of the velocity of the particle and is directly proportional to its specific charge $\dfrac{q}{m}$.

Let us compare the specific charge $\dfrac{q}{m}$ for the given particles:

For an electron, $\dfrac{q}{m} = \dfrac{e}{m_e}$

For a proton, $\dfrac{q}{m} = \dfrac{e}{m_p}$

For $\text{He}^+$, $\dfrac{q}{m} = \dfrac{e}{4m_p}$

For $\text{Li}^+$, $\dfrac{q}{m} = \dfrac{e}{7m_p}$

Since $\text{Li}^+$ has the largest mass among the given ions with the same magnitude of charge $e$, it has the minimum value of $\dfrac{q}{m}$.

Therefore, $\text{Li}^+$ will exhibit the minimum frequency of revolution.