Table 1 shows data of speed $v$ and kinetic energy $E_k$ for electrons from a modern version of the Bertozzi experiment. Table 1 | $v / 10^8 \, \mathrm{m \, s^{-1}}$ | $E_k \, \mathrm{MeV}$ | |-----------------------------------|---------------------| | 2.60 | 0.5 | | 2.73 | 0.7 | | 2.88 | 1.3 | | 2.96 | 2.6 | | 2.99 | 5.8 | Classical mechanics predicts that $E_k \propto v^2$. 1. Deduce whether the data in Table 1 is consistent with this prediction. 2. Discuss how Einstein’s theory of special relativity explains the data in Table 1. 3. Calculate, in J, the kinetic energy of one electron travelling at a speed of 0.95c.

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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

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Table 1 shows data of speed $v$ and kinetic energy $E_k$ for electrons from a modern version of the Bertozzi experiment. Table 1 | $v / 10^8 \, \mathrm{m \, s^{-1}... show full transcript

Worked Solution & Example Answer: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

Step 1

Deduce whether the data in Table 1 is consistent with this prediction.

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Answer

To determine if the data is consistent with the prediction that $E_k \propto v^2$ , we can analyze at least two data points from the table and compute the ratio of $E_k$ to $v^2$ for these values:

For $v = 2.60 \times 10^8 \, \mathrm{m \, s^{-1}}$ ,
$E_k = 0.5 \mathrm{MeV} \Rightarrow E_k = 0.5 \times 1.6 \times 10^{-13} J = 8.0 \times 10^{-14} J \quad \text{(conversion to J)}$
$v^2 = (2.60 \times 10^8)^2 \approx 6.76 \times 10^{16} \mathrm{m^2 \, s^{-2}}$
Therefore,
$\frac{E_k}{v^2} = \frac{8.0 \times 10^{-14}}{6.76 \times 10^{16}} \approx 1.18 \times 10^{-30}$
For $v = 2.73 \times 10^8 \, \mathrm{m \, s^{-1}}$ ,
$E_k = 0.7 \mathrm{MeV} \Rightarrow E_k = 0.7 \times 1.6 \times 10^{-13} J = 1.12 \times 10^{-13} J$
$v^2 = (2.73 \times 10^8)^2 \approx 7.45 \times 10^{16} \mathrm{m^2 \, s^{-2}}$
Therefore,
$\frac{E_k}{v^2} = \frac{1.12 \times 10^{-13}}{7.45 \times 10^{16}} \approx 1.50 \times 10^{-30}$

Since both calculations yield similar constants, we can conclude that the data in Table 1 is consistent with the prediction that $E_k \propto v^2$ .

Step 2

Discuss how Einstein’s theory of special relativity explains the data in Table 1.

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Answer

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.

Step 3

Calculate, in J, the kinetic energy of one electron travelling at a speed of 0.95c.

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Answer

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

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

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,
  $E_0 = m_0 c^2 = 9.11 \times 10^{-31} \times (3.00 \times 10^8)^2 \approx 8.19 \times 10^{-14} \, \mathrm{J}$
Next, calculate the Lorentz factor for $v = 0.95c$ : $\gamma = \frac{1}{\sqrt{1 - (0.95)^2}} \approx 2.86$
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 $2.36 \times 10^{-13} \, \mathrm{J}$ .

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

Worked Solution & Example Answer: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

Deduce whether the data in Table 1 is consistent with this prediction.

Discuss how Einstein’s theory of special relativity explains the data in Table 1.

Calculate, in J, the kinetic energy of one electron travelling at a speed of 0.95c.

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