Question

An evacuated tube uses an accelerating voltage of 40 kV to accelerate electrons to hit a copper plate and produce x rays. Nonrelativistically, what would be the maximum speed of these electrons

Answer 1

1.187 x 10^8 m/s

Step-by-step explanation:

the potential of the electric field V = 40 kV = 40000 V

the charge on an electron e = 1.6 x 10^-19 C

The energy of an accelerated electron in an electric field is given as

E = eV

E = 1.6 x 10^-19 x 40000 = 6.4 x 10^-15 J

This energy is equal to the kinetic energy with which the electron moves, according to the conservation of energy.

The kinetic energy =
`(1)/(2)mv^(2)`

where

m is the mass of the electron = 9.109 x 10^-31

v is the speed of the electron.

Equating the energy, we have

6.4 x 10^-15 =
`(1)/(2)*9.109*10^-31*v^(2)`

6.4 x 10^-15 = 4.55 x 10^-31
`v^(2)`

`v^(2)` = 1.41 x 10^16

`x^(2) v = \sqrt{1.41*10^(16)}` = 1.187 x 10^8 m/s