the electron beam in a television tube consists of electrons accelerated from rest through a potential difference of about 20000v.what is the speed of the electrons?(ignore relativistic effects).electron rest mass is 9.11 X 10^-31kg and electron charge is 1.6 x 10^-19
2 answers
Ke = (1/2) m v^2 = 20000 volts * electron charge
ChangePE + changeKE = 0.
the change in potential energy of electron is = dPE = qdV = (-e)dV.
The change in kinetic energy of electron is = 1/2mV^2 - 0 = 1/2mV^2.
put these together and get.
(-e)dV + 1/2mV^2 = 0.
1/2mV^2 = (e)dV.
mV^2 = 2(e)dV.
V^2 = 2edV/m.
Insert the figures
V^2 = 2(1.60 * 10^-19C)(20000V)/(9.11 * 10^-31Kg) = 7.0 * 10^15
V = ¡Ì7.0 * 10^15 = 8.4 * 10^7m/s
the change in potential energy of electron is = dPE = qdV = (-e)dV.
The change in kinetic energy of electron is = 1/2mV^2 - 0 = 1/2mV^2.
put these together and get.
(-e)dV + 1/2mV^2 = 0.
1/2mV^2 = (e)dV.
mV^2 = 2(e)dV.
V^2 = 2edV/m.
Insert the figures
V^2 = 2(1.60 * 10^-19C)(20000V)/(9.11 * 10^-31Kg) = 7.0 * 10^15
V = ¡Ì7.0 * 10^15 = 8.4 * 10^7m/s
1 answer