Calculate the speed of a proton that is accelerated from rest through an electri

Question

Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 145 V. Calculate the speed of an electron that is accelerated through the same potential difference. Part 1 of 5 - Conceptualize For this moderate potential difference, we expect that the final speed of the particles will be substantially less than the speed of light. Because of the much greater mass of the proton, we expect the speed of the electron to be significantly greater than the speed of the proton. An equal force on both particles will result in a much greater acceleration for the electron. Part 2 of 5 - Categorize We apply conservation of energy to each particle-field system to find the final speed from the kinetic energy of the particle. Part 3 of 5 - Analyze For the proton-field system, energy is conserved as the proton moves from high to low potential. We have K_i + U_i = K_f + U_f. This becomes the following conservation of energy equation. 0 + qV_i = 1/2 m_pV_p^2 + 0 the initial potential energy of the proton is qV_i = (l.60 times 10^-19 c) (_)(1j/C/1v)= 1.66 Your response differs from the correct answer by more than 10%. Double check your calculations, times 10^-17 J.

Explanation / Answer

a)

Use Work-energy theorem,

Workdone = change in kinetic energy

F*d = (1/2)*m*v^2

q*E*d = (1/2)*m*v^2

q*delta_V = (1/2)*m*v^2

v = sqrt(2*q*delta_V/m)

= sqrt(2*1.6*10^-19*145/(1.67*10^-27))

= 1.67*10^5 m/s

b) for elctron,

v = sqrt(2*q*delta_V/m)

= sqrt(2*1.6*10^-19*145/(9.1*10^-31))

= 7.14*10^6 m/s

3 of 5

a)

q*Vi = 1.6*10^-19*145 V = 2.32*10^-7 J

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