To find the maximum speed of the electrons non-relativistically, we can use the principle of energy conservation, where the kinetic energy of the electrons is equal to the electrical potential energy provided by the accelerating voltage.

Given:

• Accelerating voltage $V = 88 \, \text{kV} = 88 \times 10^3 \, \text{V}$
• Charge of an electron $e = 1.602 \times 10^{-19} \, \text{C}$
• Mass of an electron $m_e = 9.109 \times 10^{-31} \, \text{kg}$

The kinetic energy $K.E.$ gained by the electron is given by: $K.E. = eV$

This kinetic energy is also equal to the non-relativistic kinetic energy formula: $K.E. = \frac{1}{2} m_e v^2$

Equating the two expressions for kinetic energy: $eV = \frac{1}{2} m_e v^2$

Solving for the speed $v$: $v = \sqrt{\frac{2eV}{m_e}}$

Now, plug in the values: $v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \, \text{C} \times 88 \times 10^3 \, \text{V}}{9.109 \times 10^{-31} \, \text{kg}}}$

Let's calculate this:

$v = \sqrt{\frac{2 \times 1.602 \times 10^{-19} \times 88 \times 10^3}{9.109 \times 10^{-31}}}$

$v \approx \sqrt{\frac{2.816 \times 10^{-14}}{9.109 \times 10^{-31}}}$

$v \approx \sqrt{3.09 \times 10^{16}} \, \text{m/s}$

$v \approx 5.56 \times 10^8 \, \text{m/s}$

So, the maximum speed of the electrons, non-relativistically, is approximately $5.56 \times 10^8 \, \text{m/s}$.

Would you like further details or have any questions?

Here are some related questions:

1. What is the relativistic speed of the electrons with the same voltage?
2. How does the energy of X-rays relate to the speed of electrons?
3. What would be the speed if the voltage was doubled?
4. How does the electron mass affect the speed calculation?
5. What is the significance of using copper in X-ray tubes?

Tip: Non-relativistic calculations are accurate only when the velocity is significantly less than the speed of light; otherwise, relativistic effects must be considered.