An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V$_e$ and V$_p$ respectively. Which one of the following relations is correct?

When a charged particle is accelerated from rest through a potential difference V, its kinetic energy is equal to the work done on it by the electric field, which is |qV|. So, (1/2)mv^2 = |qV|. The speed is given by v = sqrt(2|q|V/m). The electron and the proton have the same magnitude of charge (|q| = e) and are accelerated through the same potential difference V. Thus, their kinetic energies gained are equal: KE_e = KE_p = eV. (1/2)m_e V_e^2 = (1/2)m_p V_p^2 m_e V_e^2 = m_p V_p^2 V_e / V_p = sqrt(m_p / m_e) Since the mass of the electron (m_e) is significantly less than the mass of the proton (m_p), the ratio m_p / m_e is greater than 1. Therefore, V_e / V_p > 1, which means V_e > V_p. The lighter particle (electron) achieves a much higher speed than the heavier particle (proton) for the same kinetic energy.