A doubly-ionized lithium ion (m = 7, 1 u = 1.66E-27 kg) is accelerated from rest

Question

A doubly-ionized lithium ion (m = 7, 1 u = 1.66E-27 kg) is accelerated from rest through a uniform electric field. The ion travels 0.500m in the field and emerges from the filed witha speed of 4.30E5 m/s. a. Calculate the ion's kinetic energy inelectron-Volts. b. What is the change in the electric potentialexperiences by the ion? c. What was the strength of the electricfield? d. How long did this acceleration process takeplace? A doubly-ionized lithium ion (m = 7, 1 u = 1.66E-27 kg) is accelerated from rest through a uniform electric field. The ion travels 0.500m in the field and emerges from the filed witha speed of 4.30E5 m/s. a. Calculate the ion's kinetic energy inelectron-Volts. b. What is the change in the electric potentialexperiences by the ion? c. What was the strength of the electricfield? d. How long did this acceleration process takeplace?

Explanation / Answer

KE = 2 qe V = 1/2 m v2 KE = 1/2 * 7 * 1.66 * 10E-27 * (4.3 * 10E5)2 = 1.07* 10E-15 J KE = 1/2 * 7 * 1.66 * 10E-27 * (4.3 * 10E5)2 = 1.07* 10E-15 J KE (eV) = 1.07 * 10E-15 / 1.6 * 10E-19 = 6.71 * 10E3 eV V = E d or V = W / Q = 1.07 * 10E-15 / (2 *1.6 * 10E-19) = 3.34 * 10E3 V    (change inpotential) E = V / d = 3.34 * 10E3 / .5 = 6.68 * 10E3 N / C s = 1/2 a t2   where a = F / m = E Q / m= 2 E e / m t = (s m / (E e)) = (.5 * 7 * 1.66 * 10E-27 /(6.68 * 10E3 * 1.6 * 10E-19)) = 2.33 * 10E-6 sec  

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