Table 1 shows data of speed $v$ and kinetic energy $E_k$ for electrons from a modern version of the Bertozzi experiment - AQA - A-Level Physics - Question 4 - 2019 - Paper 7

Einstein's theory of special relativity introduces the concept of relativistic mass, which changes as an object approaches the speed of light, $c$. As velocity increases, so does the kinetic energy, following the relation:

$E_k = \gamma m_0 c^2 - m_0 c^2$

where $\gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}}$ is the Lorentz factor and $m_0$ is the rest mass. For low speeds, this leads to the classical approximation $E_k \approx \frac{1}{2} m v^2$, but as $v$ approaches $c$, the increase in kinetic energy becomes nonlinear.

In the data from Table 1, we see kinetic energies that don’t align with classical predictions as the speeds increase. For example, at $v = 2.99 \times 10^8 \, \mathrm{m \, s^{-1}}$, the kinetic energy vastly exceeds what would be predicted classically, demonstrating the need for relativistic considerations. Einstein's theory effectively accounts for the differences, illustrating that at high velocities, mass and energy behave in ways that classical mechanics cannot accurately describe.

To calculate the kinetic energy of one electron travelling at 0.95c, we again use the relation from Einstein's theory:

$E_k = \gamma m_0 c^2 - m_0 c^2$

1. First, calculate the rest mass energy of the electron:

   • The rest mass $m_0$ of an electron is approximately $9.11 \times 10^{-31} \, \mathrm{kg}$. Thus,
     $E_0 = m_0 c^2 = 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2 \approx 8.19 \times 10^{-14} \, \mathrm{J}$

2. Next, calculate the Lorentz factor for $v = 0.95c$: $\gamma = \frac{1}{\sqrt{1 - (0.95)^2}} \approx 2.86$

3. Finally, using these values: $E_k = \gamma m_0 c^2 - m_0 c^2 = (2.86 \times 8.19 \times 10^{-14}) - (8.19 \times 10^{-14})$
   $E_k \approx 2.36 \times 10^{-13} \, \mathrm{J}$

Thus, the kinetic energy of one electron travelling at a speed of 0.95c is approximately $2.36 \times 10^{-13} \, \mathrm{J}$.