A doubly-ionized lithium ion (m = 7u, 1u=1.66 x10 -27 kg) is accelerated from re

Question

A doubly-ionized lithium ion (m = 7u, 1u=1.66 x10-27 kg) is accelerated from rest through a uniformelectric field. The ion travels 0.500m in the field and emergesfrom the field with a speed of 8.60x105 m/s. a) Calculate the ion's kinetic energy in electron-Volts. b) What is the change in the electric potential experienced bythe ion? c) How long did this acceleration process take place? d) What was the strength of the electric field? A doubly-ionized lithium ion (m = 7u, 1u=1.66 x10-27 kg) is accelerated from rest through a uniformelectric field. The ion travels 0.500m in the field and emergesfrom the field with a speed of 8.60x105 m/s. a) Calculate the ion's kinetic energy in electron-Volts. b) What is the change in the electric potential experienced bythe ion? c) How long did this acceleration process take place? d) What was the strength of the electric field?

Explanation / Answer

a) Formula for the calculating the value of the ion's kineticenergy is               K = (1/2)mv2                   = (1/2)(7)(1.66 x 10-27 kg)(8.60x105m/s)2                   = 429.7*10-17J Since we have 1eV =1.6*10-19J      Then   K = (429.7*10-17) /(1.6*10-19)                  = 268.6*102 eV   b) We know that K = qV                           = (+2e)(V)                    V = K / (2e)                          = (429.7*10-17J) /[2(1.6*10-19C)]   d) Electric field strength is E = V / d Here V is substituted here which is calculated from thepart (b)           d =0.500m c) Given that intial speed (u) of the ion = 0 Now we know that formula                  v = u + at                     =0+at                    = at                    = (Eq / m)t                    = (2eE / m)t                 t = (mv) / (2eE) Here m = 7(1.66 x 10-27 kg)            v =8.60x105 m/s          e=  1.6*10-19C          E issubstituted here which is calculated from the part(d)                      = 429.7*10-17J Since we have 1eV =1.6*10-19J      Then   K = (429.7*10-17) /(1.6*10-19)                  = 268.6*102 eV   b) We know that K = qV                           = (+2e)(V)                    V = K / (2e)                          = (429.7*10-17J) /[2(1.6*10-19C)]   d) Electric field strength is E = V / d Here V is substituted here which is calculated from thepart (b)           d =0.500m c) Given that intial speed (u) of the ion = 0 Now we know that formula                  v = u + at                     =0+at                    = at                    = (Eq / m)t                    = (2eE / m)t                 t = (mv) / (2eE) Here m = 7(1.66 x 10-27 kg)            v =8.60x105 m/s          e=  1.6*10-19C          E issubstituted here which is calculated from the part(d)   

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