Given
that the ground state hydrogen electron has a kinetic energy of 13.6
eV, and a velocity of 1/137 th the speed of light; what is the velocity
of an electron after being accelerated by a 54.4 Volt electric field?
(Hint: 54.4/13.6 = 4)
Ans: (2/137) c, or 0.0146 c
After
dropping through a 54.4 V field, the electron has four times the
kinetic energy of the hydrogen electron. Since kinetic energy is
proportional to velocity squared, the velocity is only twice the
velocity of the hydrogen electron. |