A Van de Graaff accelerator utilizes a \(50.0 \mathrm{MV}\) potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?

First, we need to compute the energy gained by the proton and electron when they are accelerated by the potential difference of \(50.0 MV\). Given potential difference: \(V = 50.0 \times 10^6 V\) Charge of a proton: \(q_p = +e = +1.60 \times 10^{-19} C\) Charge of an electron: \(q_e = -e = -1.60 \times 10^{-19} C\) Using the potential energy formula, we have: Energy gained by proton \(= U_p = q_p V = (+1.60 \times 10^{-19}C)(50.0 \times 10^6 V) = 8.00 \times 10^{-12} J\) Energy gained by the electron \(= U_e = q_e V = (-1.60 \times 10^{-19}C)(50.0 \times 10^6 V) = -8.00 \times 10^{-12} J\) Since both particles are moved in the direction of the electric field, they gain positive energy.

Next, we need to calculate the final velocities of the proton and electron after gaining the computed energies. Mass of a proton: \(m_p = 1.67 \times 10^{-27}kg\) Mass of an electron: \(m_e = 9.11 \times 10^{-31}kg\) Using the kinetic energy formula and solving for the velocity, we have: Proton's velocity: \(v_p = \sqrt{\dfrac{2K_p}{m_p}}\) = \(\sqrt{\dfrac{2(8.00 \times 10^{-12}J)}{1.67 \times 10^{-27}kg}} = 4.37 \times 10^6 m/s\) Electron's velocity: \(v_e = \sqrt{\dfrac{2K_e}{m_e}}\) = \(\sqrt{\dfrac{2(8.00 \times 10^{-12}J)}{9.11 \times 10^{-31}kg}} = 1.326 \times 10^8 m/s\) Thus, the velocities of the proton and the electron after getting accelerated by the potential difference of \(50.0 MV\) are approximately \(4.37 \times 10^6 m/s\) and \(1.326 \times 10^8 m/s\), respectively.