A proton, an electron and an \(\alpha\)-particle enter at right angles to a uniform magnetic field with the same velocity. If \(R_p, R_e\) and \(R_\alpha\) are the radii of circular paths of these particles, then

The radius of the circular path of a charged particle in a uniform magnetic field is given by \(R = \dfrac{mv}{qB}\).

Since the velocity \(v\) and magnetic field \(B\) are the same for all particles, \(R \propto \dfrac{m}{q}\).

For a proton: \(R_p \propto \dfrac{m_p}{e}\)

For an electron: \(R_e \propto \dfrac{m_e}{e}\)

For an \(\alpha\)-particle: \(R_\alpha \propto \dfrac{m_\alpha}{q_\alpha} = \dfrac{4m_p}{2e} = \dfrac{2m_p}{e}\)

Since \(m_e < m_p\), we have \(\dfrac{m_e}{e} < \dfrac{m_p}{e} < \dfrac{2m_p}{e}\).

Therefore, \(R_e < R_p < R_\alpha\), which can be written as \(R_\alpha > R_p > R_e\).

Answer: \(R_\alpha > R_p > R_e\)