Answer to Question #138553 in Electric Circuits for Kaila

Question

Find the electric potential difference (change in Ve) required to stop an electron (called a “stopping potential”) moving with an initial speed of 2.85 x 10^7 m/s. Would a proton travelling at the same speed require a greater or lesser magnitude of electric potential difference? Explain.

Answer 1

Explanations& Calculations

• By the application of a potential difference to the electron, work is done on it in the opposite direction with an amount equal to the energy it already has: the kinetic energy.
• Work done is $\small W = q\Delta V$
• And, $\qquad\qquad \begin{aligned} \small W &= \small \frac{1}{2}m_ev^2\,\,\,\,\,:m_e = 9.109\times 10^{-31}kg\\ \small \Delta V &= \small \frac{m_ev^2}{2q}\\ &= \small \frac{9.109\times 10^{-31}kg\times (2.85\times 10^7m/s)^2}{2\times(1.6\times10^{-19}C)}\\ &= \small \bold{2312.12V=2.312\,MV} \end{aligned}$

• Even though the speed is the same, the mass of a proton is larger ($\small m_p =1.673 \times 10^{-27}kg$ ) than that of an electron meaning that it would need a greater work done due the higher kinetic energy it has. This explains a need of a higher stopping potential.