What is the change in kinetic energy of an al | vedclass

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(A) The change in kinetic energy $\Delta K$ of a charged particle moving in an electric field is given by $\Delta K = -\Delta U = -q\Delta V = q(V_i -

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$40\ eV$

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(A) The change in kinetic energy $\Delta K$ of a charged particle moving in an electric field is given by $\Delta K = -\Delta U = -q\Delta V = q(V_i - V_f)$.For an $\alpha$-particle,the charge $q = +2e$.The potential difference is $\Delta V = V_f - V_i = 50\ V - 70\ V = -20\ V$.Therefore,the change in kinetic energy is $\Delta K = -q\Delta V = -(2e) 	imes (-20\ V) = 40\ eV$.Since the particle moves from a higher potential to a lower potential,its kinetic energy increases by $40\ eV$.

What is the change in kinetic energy of an $\alpha$-particle when it moves from a point at $70\ V$ to a point at $50\ V$?

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