(a) Calculate the speed of a proton that is accelerated from rest through a potential difference of 111 V.?

Question

(b) Calculate the speed of an electron that is accelerated through the same potential difference.

Answer 1

Try using qV = KE = 0.2mv^2

q = charge of the particle

v = potential difference

KE = equation for kinetic energy

Answer 2

potential distinction = paintings accomplished consistent with unit value consequently paintings accomplished = potential distinction * value value on a proton = a million.6 * 10^-19 C potential distinction = 142 V consequently paintings W = 142 * a million.6 * 10^-19 J = 227.2 * 10^-19 J = 2.27 * 10^-17 J enable velocity of proton = v preliminary kinetic power = 0 because of the fact the proton is initially at relax. very final kinetic power = a million/2 mv^2 the place m = mass of proton substitute in ok.E. = a million/2 mv^2 - 0 = a million/2 mv^2 via paintings power theorem substitute in ok.E. = paintings accomplished Or a million/2 mv^2 = W Or v = sqrt(2W/m)------------------------------... Or v = sqrt{2 * 2.27 * 10^-17/(a million.sixty seven * 10^-27)} = sqrt{(2 * 2.27/a million.sixty seven) * 10^(-17 + 27)} = sqrt(2.seventy two * 10^10) = a million.sixty 5 * 10^5 m/s paintings accomplished = potential distinction * value P.E. is the comparable i.e. 142 eV. the two proton and electron have the comparable value of value (purely sign is distinctive). consequently, value of the paintings accomplished is likewise the comparable. consequently W = 2.27 * 10^-17 J as calculated formerly In equation (a million), placed the mass of the electron in place of m. m = 9.a million * 10^-31 kg v = sqrt(2W/m) Or v = sqrt{2 * 2.27 * 10^-17/(9.a million * 10^-31)} = sqrt{(2 * 2.27/9.a million) * 10^(-17 + 31)} = sqrt(0.5 * 10^14) = sqrt(50 * 10^12) = 7.a million * 10^6 m/s Ans: velocity of proton = a million.sixty 5 * 10^5 m/s velocity of electron = 7.a million * 10^6 m/s

Answer 3

0.5*m*velocity^2 = charge*potential

u need to know electron's mass , to get velocity