Question

7. In the Bohr model of atomic hydrogen, an electron of mass 9.11 x 10-31 kg revolves around the proton in a circular orbit of radius .29 x 10-11 m. The proton and electron have equal charge magnitudes and the mass of the proton is 1.67 x10-27 kg. What is the radial acceleration of the electron? What is it's velocity? What is it's angular velocity?

Answer 1

Electron is moving on a circular path. The electrostatic force b/w proton & electron provide centripetal force Electric force = Centripetal force 1/4pi epsilon = q_r q_e/r^2 = m_e v^2/r equation (I) Radial acceleration a_r = v^2/r equation (II) from eqn (I) 1/4pi epsilon q_r q_e/r^2 = m_e a_r a_r = 1/4pi epsilon 1/m q_p q_e/r^2 equation(III) \r\n We have, 1/4pi epsilon = 9 times 10^9 Nm^2/c^2 Charge on electron/proton Vertical-bar q_p Vertical-bar = q_e = 1.6 times 10^-19 C Radius r = 0.25 times 10^-11 m Mass of Electron m_e = 9.1 times 10^-31 Kg Then plugging in these values in eqn (III) a_r = (9 times 10^9 Nm^2/C^2) (1.6 times 10^-15)^2/9.1 times 10^-31 times (0.25 times 10^-11 m)^2 a_r = 3.01 times 10^25 m/s^2 Next Radial acceleration a_r = v^2/r Speed v = squareroot a_r times r v = squareroot 3.0 times 10^25 m/s^2 times 0.29 times 10^-11 m v = 9.34 times 10^6 m/s Next Angular Velocity, w =