Table 1 shows data of speed $v$ and kinetic energy $E_k$ for electrons from a modern version of the Bertozzi experiment. Table 1 | $v / 10^8 \text{ m s}^{-1}$ | $E_k / \text{ MeV}$ | |------------------------------|---------------------| | 2.60 | 0.5 | | 2.73 | 0.7 | | 2.88 | 1.3 | | 2.96 | 2.6 | | 2.99 | 5.8 | 0 4 . 1 Classical mechanics predicts that $E_k \propto v^2$. Deduce whether the data in Table 1 are consistent with this prediction. 0 4 . 2 Discuss how Einstein’s theory of special relativity explains the data in Table 1. 0 4 . 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 | $v / 10^8 \text{ m s}^{-1}$ | $E... 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 are consistent with this prediction.

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Answer

To determine if the data in Table 1 are consistent with the prediction that $E_k \propto v^2$ , we can calculate the ratio of $E_k$ for different pairs of speeds.

Select data sets: For example, using the pairs (2.60, 0.5) and (2.99, 5.8).
Calculate $v^2$ for each pair:
- For speed $v = 2.60 \times 10^8 \text{ m s}^{-1}$ : $v^2 = (2.60 \times 10^8)^2 = 6.76 \times 10^{16} \text{ m}^2 \text{s}^{-2}$
- For speed $v = 2.99 \times 10^8 \text{ m s}^{-1}$ : $v^2 = (2.99 \times 10^8)^2 = 8.94 \times 10^{16} \text{ m}^2 \text{s}^{-2}$
Calculate the ratio of $E_k$ values:
- For $E_k = 0.5$ MeV at 2.60:
- For $E_k = 5.8$ MeV at 2.99:
- Ratio is approximately: $\frac{5.8 \text{ MeV}}{0.5 \text{ MeV}} = 11.6$
- The ratio of the square speeds is: $\frac{8.94 \times 10^{16}}{6.76 \times 10^{16}} \approx 1.32$

Since the ratios suggest an increase consistent with $E_k \propto v^2$ , the data appears consistent with classical predictions.

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 addresses the behavior of particles at relativistic speeds. Key points include:

Mass Increase: As electrons approach the speed of light, their relativistic mass increases, which means that: $E_k = (\gamma - 1) mc^2$ where $\, \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$ . This indicates that kinetic energy increases more steeply than predicted by classical mechanics, as speed increases towards $c$ .
Kinetic Energy Behavior: The data shows a nonlinear relationship between speed and kinetic energy, as evidenced by the rapid increase of $E_k$ at higher speeds in Table 1. This indicates that as speed approaches light speed ( $c$ ), there is a significant increase in energy.
Conclusion: The greater than $v^2$ relationship observed in Table 1 supports the prediction of Einstein’s theory, which states that mass-energy equivalence significantly alters the classical interpretation of kinetic energy at high speeds.

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 an electron moving at a speed of $0.95c$ , we use:

Mass of Electron: The rest mass $m = 9.11 \times 10^{-31} \text{ kg}$ .
Speed: Let $v = 0.95c$ , where $c = 3.00 \times 10^8 \text{ m s}^{-1}$ : $v = 0.95 \times 3.00 \times 10^8 \approx 2.85 \times 10^8 \text{ m s}^{-1}$
Calculate $\, \gamma$ : $\gamma = \frac{1}{\sqrt{1 - \frac{(0.95c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.9025}} = \frac{1}{\sqrt{0.0975}} \approx 3.19$
Kinetic Energy Calculation: Using the formula for kinetic energy: $E_k = (\gamma - 1) mc^2$ So: $E_k = (3.19 - 1)(9.11 \times 10^{-31})(3.00 \times 10^8)^2$ $E_k = 2.19 imes (9.11 \times 10^{-31})(9.00 \times 10^{16})\approx 2.00 \times 10^{-14} \text{ J}$

Therefore, the kinetic energy of one electron travelling at a speed of $0.95c$ is approximately $2.00 \times 10^{-14} \text{ J}$ .

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 are 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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