Question

An evacuated tube uses a potential difference of ΔV = 0.44 kV to accelerate electrons, which then hit a copper plate and produce X-rays.

A. Write an expression for the non-relativistic speed of these electrons v in terms of e, ΔV, and m, assuming the electrons start from rest.
B. Calculate the non-relativistic speed of these electrons v in m/s.

Answer 1

To find the non-relativistic speed of electrons accelerated by a potential difference of 0.44 kV, the energy principle is applied giving the formula v = sqrt((2qΔV)/m). By plugging in the values for the electron's charge and mass and the potential difference, the speed is calculated to be approximately 1.24 × 10^7 m/s.

Step-by-step explanation:

To determine the non-relativistic speed v of electrons accelerated across a potential difference ΔV = 0.44 kV, we can use the work-energy principle in physics. According to this principle, the work done on the electrons by the electric field as they move through the potential difference is equal to their gain in kinetic energy:

Work done by electric field = Kinetic energy of electrons

qΔV = (1/2)m v2

Here, q is the charge of an electron (1.60 × 10-19 C), and m is the mass of an electron (9.11 × 10-31 kg). Rearranging for v:

v = sqrt((2qΔV)/m)

Let's substitute the given values and calculate the speed:

v = sqrt((2 * 1.60 × 10-19 C * 0.44 × 103) / (9.11 × 10-31 kg))

v = sqrt((2 * 1.60 × 10-19 C * 440 V) / (9.11 × 10-31 kg))

v ≈ 1.24 × 107 m/s

So, the non-relativistic speed of the electrons accelerated by a potential difference of 0.44 kV is approximately 1.24 × 107 m/s.