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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 are consistent with this prediction.

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Answer

To determine if the data support the prediction that kinetic energy Ekv2 E_k \propto v^2, we will consider the ratio of EkE_k to v2v^2 for at least two sets of data from Table 1.

Calculating:

  • For v=2.60×108m s1v = 2.60 \times 10^8 \mathrm{m \ s^{-1}}:

    • Ek=0.5MeV=0.5×1.6×1013J8.0×1014JE_k = 0.5 \mathrm{MeV} = 0.5 \times 1.6 \times 10^{-13} \mathrm{J} \approx 8.0 \times 10^{-14} \mathrm{J}.
    • v2=(2.60×108)26.76×1016m2 s2v^2 = (2.60 \times 10^8)^2 \approx 6.76 \times 10^{16} \mathrm{m^2 \ s^{-2}}.
    • Thus, the ratio rac{E_k}{v^2} \approx \frac{8.0 \times 10^{-14}}{6.76 \times 10^{16}} \approx 1.18 \times 10^{-30}.
  • For v=2.99×108m s1v = 2.99 \times 10^8 \mathrm{m \ s^{-1}}:

    • Ek=5.8MeV9.28×1013JE_k = 5.8 \mathrm{MeV} \approx 9.28 \times 10^{-13} \mathrm{J}.
    • v2=(2.99×108)28.94×1016m2 s2v^2 = (2.99 \times 10^8)^2 \approx 8.94 \times 10^{16} \mathrm{m^2 \ s^{-2}}.
    • Thus, the ratio rac{E_k}{v^2} \approx \frac{9.28 \times 10^{-13}}{8.94 \times 10^{16}} \approx 1.04 \times 10^{-29}.

Since these ratios decrease with increasing vv, the data suggest that the relationship between EkE_k and v2v^2 may not be direct due to relativistic effects as speeds approach the speed of light.

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 posits that as particles like electrons accelerate to speeds close to the speed of light, their mass effectively increases, which in turn affects their kinetic energy.

As per special relativity, the relativistic kinetic energy EkE_k can be described by the formula: Ek=(γ1)m0c2E_k = (\gamma - 1) m_0 c^2 where γ=11(v2/c2) \gamma = \frac{1}{\sqrt{1 - (v^2 / c^2)}} and m0m_0 is the rest mass of the electron.

As the velocity vv of the electron approaches the speed of light cc, γ \gamma increases significantly, and thus the kinetic energy EkE_k increases disproportionately compared to classical predictions. This explains why the data in Table 1 shows kinetic energy that does not align with the classical v2v^2 relationship, instead indicating a more complex relationship dictated by relativistic effects.

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 for an electron travelling at a speed of 0.95c0.95c, we first find γ \gamma: γ=11(0.95)22.83.\gamma = \frac{1}{\sqrt{1 - (0.95)^2}} \approx 2.83.

Using the rest mass of the electron, m09.11×1031kgm_0 \approx 9.11 \times 10^{-31} \mathrm{kg}, the kinetic energy is given by: Ek=(γ1)m0c2E_k = (\gamma - 1) m_0 c^2 Substituting the known values, Ek=(2.831)(9.11×1031kg)(3.00×108m s1)2E_k = (2.83 - 1)(9.11 \times 10^{-31} \mathrm{kg})(3.00 \times 10^8 \mathrm{m \ s^{-1}})^2 Ek=1.83×(9.11×1031)(9.00×1016)1.50×1014extJ.E_k = 1.83 \times (9.11 \times 10^{-31})(9.00 \times 10^{16})\approx 1.50 \times 10^{-14} ext{ J}.

Thus, the kinetic energy of one electron travelling at a speed of 0.95c is approximately 1.50×10141.50 \times 10^{-14} J.

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