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An electron with an initial kinetic energy of 71.17 eV is projected into a region of...

An electron with an initial kinetic energy of 71.17 eV is projected into a region of constant electric field of 1*10^5 V/m such that the electron decelerates . What distance does the electron travel before it stops and starts to travel the opposite direction?

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Answer #1

KE = 71.17*1.6*10^-19


0.5*m*u^2 = 1.139*10^-17


0.5*9.1*10^-31*u^2 = 1.139*10^-17


u = sqrt(2*1.139*10^-17/9.1*10^-31)


= 5*10^6 m/s (initial speedd)


Fe = q*E


m*a = q*E


a = qE/m


= 1.6*10^-19*10^5/9.1*10^-31


= -1.76*10^16 m/s^2 (acceleartino)


we know, v^2 - u^2 = 2*a*s


here, v = 0


==> s = -u^2/(2*a)


= -(5*10^6)^2/(-2*1.76*10^16)


= 7.1*10^-4 m


= 0.71 mm


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Answer #2

let the electron travel x m before its velocity becomes 0.

so work done by electric field = initial KE of electron

initial KE = 71.17 eV = 71.17 *(1.6*10^-19) J

work done by electric field = (1*10^5)*(1.6*10^-19)* x


so   (1*10^5)*(1.6*10^-19)* x = 71.17 *(1.6*10^-19)

=> x = 7.117*10^-4 m

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