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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} \quad... 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 relationship Ekv2E_k \propto v^2, we can calculate the ratio of EkE_k to v2v^2 using at least two data points from the table.

Take the first two data sets:

  • For v=2.60×108m s1v = 2.60 \times 10^8 \, \text{m s}^{-1}, Ek=0.5MeVE_k = 0.5 \text{MeV}:

    • Calculate v2=(2.60×108)26.76×1016v^2 = (2.60 \times 10^8)^2 \approx 6.76 \times 10^{16},
    • Calculate the ratio Ekv2=0.56.76×10167.4×1018\frac{E_k}{v^2} = \frac{0.5}{6.76 \times 10^{16}} \approx 7.4 \times 10^{-18}.
  • For v=2.73×108m s1v = 2.73 \times 10^8 \, \text{m s}^{-1}, Ek=0.7MeVE_k = 0.7 \text{MeV}:

    • Calculate v27.45×1016v^2 \approx 7.45 \times 10^{16},
    • Calculate the ratio Ekv2=0.77.45×10169.4×1018\frac{E_k}{v^2} = \frac{0.7}{7.45 \times 10^{16}} \approx 9.4 \times 10^{-18}.

From these calculations, we observe that Ekv2\frac{E_k}{v^2} is roughly constant, suggesting that the data is consistent with the prediction that Ekv2E_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 enhances our understanding of the relationship between kinetic energy and speed at relativistic speeds. As an electron approaches the speed of light, the classical formula for kinetic energy Ek=12mv2E_k = \frac{1}{2} mv^2 becomes inadequate. Instead, the relativistic kinetic energy is given by:

Ek=(γ1)mc2E_k = (\gamma - 1) mc^2

where γ=11(v2/c2)\gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}} is the Lorentz factor.

Using this relationship, as vv increases toward cc, γ\gamma increases dramatically, leading to a significant increase in kinetic energy. This explains the non-linear increases in EkE_k observed in Table 1, as speeds approach the light speed limit, indicating that mass effectively increases, resulting in a more substantial increase in energy than predicted by classical mechanics.

Step 3

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

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Answer

First, we calculate the Lorentz factor γ\gamma for v=0.95cv = 0.95c:

γ=11(0.95)22.87.\gamma = \frac{1}{\sqrt{1 - (0.95)^2}} \approx 2.87.

Using the mass of an electron me9.11×1031kgm_e \approx 9.11 \times 10^{-31} \, \text{kg} and the speed of light c3.00×108m s1c \approx 3.00 \times 10^{8} \, \text{m s}^{-1}, we can calculate the relativistic kinetic energy:

Ek=(γ1)mec2(2.871)(9.11×1031kg)(3.00×108m s1)2E_k = (\gamma - 1) m_e c^2 \approx (2.87 - 1)(9.11 \times 10^{-31} \, \text{kg})(3.00 \times 10^8 \, \text{m s}^{-1})^2

Calculating this gives:

Ek1.70×1013J.E_k \approx 1.70 \times 10^{-13} \, \text{J}.

Therefore, the kinetic energy of one electron travelling at a speed of 0.95c is approximately 1.70×1013J1.70 \times 10^{-13} \, \text{J}.

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