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(A) The work done by the electric field on the electron is equal to the change in its kinetic energy.Work done $W = q \Delta V = e(V_2 - V_1)$.Given $

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$2.65 	imes 10^{6}\ m/s$

Vedclass · Answer

(A) The work done by the electric field on the electron is equal to the change in its kinetic energy.Work done $W = q \Delta V = e(V_2 - V_1)$.Given $V_2 - V_1 = 20\ V$ and $e = 1.6 	imes 10^{-19}\ C$.So,$W = 1.6 	imes 10^{-19} 	imes 20 = 3.2 	imes 10^{-18}\ J$.According to the work-energy theorem,$W = \Delta K = \frac{1}{2} m_0 v^2 - 0$.$3.2 	imes 10^{-18} = \frac{1}{2} 	imes 9.11 	imes 10^{-31} 	imes v^2$.$v^2 = \frac{2 	imes 3.2 	imes 10^{-18}}{9.11 	imes 10^{-31}} = \frac{6.4}{9.11} 	imes 10^{13} \approx 0.7025 	imes 10^{13} = 7.025 	imes 10^{12}$.$v = \sqrt{7.025 	imes 10^{12}} \approx 2.65 	imes 10^{6}\ m/s$.

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