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Table 1 shows results of an experiment to investigate how the de Broglie wavelength λ of an electron varies with its velocity v - AQA - A-Level Physics - Question 2 - 2018 - Paper 1

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Table 1 shows results of an experiment to investigate how the de Broglie wavelength λ of an electron varies with its velocity v. | v / 10^7 m s^-1 | λ / 10^-11 ... show full transcript

Worked Solution & Example Answer:Table 1 shows results of an experiment to investigate how the de Broglie wavelength λ of an electron varies with its velocity v - AQA - A-Level Physics - Question 2 - 2018 - Paper 1

Step 1

Show that the data in Table 1 are consistent with the relationship $\lambda \propto \frac{1}{v}$

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Answer

To show that the data in Table 1 are consistent with the relationship λ1v\lambda \propto \frac{1}{v}, we can calculate the ratio of λ\lambda to vv for each pair of data values. This relationship implies that if we plot λ\lambda against 1v\frac{1}{v}, we should expect a linear relationship.

Calculating the reciprocal of vv:

  • For v=1.5×107 m/sv = 1.5 \times 10^7 \text{ m/s}, λ=4.9×1011 m\lambda = 4.9 \times 10^{-11} \text{ m}, we find: 1v=11.5×1076.67×108 m/s1 \frac{1}{v} = \frac{1}{1.5 \times 10^7} \approx 6.67 \times 10^{-8} \text{ m/s}^{-1}
  • For v=2.5×107 m/sv = 2.5 \times 10^7 \text{ m/s}, λ=2.9×1011 m\lambda = 2.9 \times 10^{-11} \text{ m}, we find: 1v4.00×108 m/s1 \frac{1}{v} \approx 4.00 \times 10^{-8} \text{ m/s}^{-1}
  • For v=3.5×107 m/sv = 3.5 \times 10^7 \text{ m/s}, λ=2.1×1011 m\lambda = 2.1 \times 10^{-11} \text{ m}, we find: 1v2.86×108 m/s1 \frac{1}{v} \approx 2.86 \times 10^{-8} \text{ m/s}^{-1}

We can see that as vv increases, λ\lambda decreases, confirming that they are inversely proportional.

Step 2

Calculate a value for the Planck constant suggested by the data in Table 1.

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Answer

Using the de Broglie wavelength formula:

λ=hmv\lambda = \frac{h}{mv}

Where:

  • hh is the Planck constant,
  • mm is the mass of the electron (approximately 9.11×1031 kg9.11 \times 10^{-31} \text{ kg}),
  • vv is the velocity.

Rearranging gives:

h=λmvh = \lambda mv

Using the first set of values (λ=4.9×1011 m\lambda = 4.9 \times 10^{-11} \text{ m} and v=1.5×107 m/sv = 1.5 \times 10^7 \text{ m/s}):

h=(4.9×1011)×(9.11×1031)×(1.5×107)7.1×1034 J sh = (4.9 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (1.5 \times 10^7) \approx 7.1 \times 10^{-34} \text{ J s}

Repeating for the other data points and averaging them may yield a more reliable value for the Planck constant.

Step 3

Explain how the pattern produced on the screen supports the idea that the electron beam is behaving as a wave rather than as a stream of particles.

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Answer

The pattern produced on the screen consists of a series of bright and dark fringes. This diffraction pattern is indicative of wave behavior, similar to what is observed in classical wave experiments, such as light diffraction through slits. The constructive interference of the electron waves leads to the bright spots while destructive interference creates dark regions.

If the electrons were behaving purely as particles, we would expect to see a straightforward impact on the screen corresponding with the graphite target, with no interference pattern. Thus, the presence of the fringes confirms the wave-like nature of electrons.

Step 4

Explain how the emission of light from the fluorescent screen shows that the electrons incident on it are behaving as particles.

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Answer

The emission of light from the fluorescent screen occurs when the electrons collide with the phosphorescent material of the screen. Each electron is a discrete particle that transfers energy to the atoms in the screen.

When these atoms are excited, they emit photons as they return to their ground state, producing visible light. This process indicates that electrons, while demonstrating wave-like behavior in their diffraction pattern, also behave as particles during interactions with matter, leading to quantized and localized emission of light.

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