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Table 1 shows results of an experiment to investigate how the de Broglie wavelength $oldsymbol{ ext{ extlambda}}$ of an electron varies with its velocity $oldsymbol{v}$ - 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 $oldsymbol{ ext{ extlambda}}$ of an electron varies with its velocity $oldsymbo... show full transcript

Worked Solution & Example Answer:Table 1 shows results of an experiment to investigate how the de Broglie wavelength $oldsymbol{ ext{ extlambda}}$ of an electron varies with its velocity $oldsymbol{v}$ - 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 demonstrate the relationship λ∝1v\lambda \propto \frac{1}{v}, we can calculate the ratio of wavelength to velocity for each entry in Table 1:

For v=1.5×107 m s−1,λ=4.9×10−11 mv = 1.5 \times 10^{7} \text{ m s}^{-1}, \lambda = 4.9 \times 10^{-11} \text{ m}: λv=4.9×10−111.5×107≈3.27×10−18\frac{\lambda}{v} = \frac{4.9 \times 10^{-11}}{1.5 \times 10^{7}} \approx 3.27 \times 10^{-18}

For v=2.5×107 m s−1,λ=2.9×10−11 mv = 2.5 \times 10^{7} \text{ m s}^{-1}, \lambda = 2.9 \times 10^{-11} \text{ m}: λv=2.9×10−112.5×107≈1.16×10−18\frac{\lambda}{v} = \frac{2.9 \times 10^{-11}}{2.5 \times 10^{7}} \approx 1.16 \times 10^{-18}

For v=3.5×107 m s−1,λ=2.1×10−11 mv = 3.5 \times 10^{7} \text{ m s}^{-1}, \lambda = 2.1 \times 10^{-11} \text{ m}: λv=2.1×10−113.5×107≈6.00×10−19\frac{\lambda}{v} = \frac{2.1 \times 10^{-11}}{3.5 \times 10^{7}} \approx 6.00 \times 10^{-19}

The ratios show that as vv increases, λ\lambda decreases, which is consistent with the inverse relationship λ∝1v\lambda \propto \frac{1}{v}.

Step 2

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

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Answer

The de Broglie wavelength is given by the formula: λ=hmv\lambda = \frac{h}{mv} where hh is the Planck constant and mm is the mass of the electron (approximately 9.11×10−31 kg9.11 \times 10^{-31} \text{ kg}). Rearranging gives: h=λmvh = \lambda mv

Substituting values from Table 1:

  1. For v=1.5×107 m s−1v = 1.5 \times 10^{7} \text{ m s}^{-1}, λ=4.9×10−11 m\lambda = 4.9 \times 10^{-11} \text{ m}: h=(4.9×10−11)×(9.11×10−31)×(1.5×107)≈7.16×10−48 J sh = (4.9 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (1.5 \times 10^{7}) \approx 7.16 \times 10^{-48} \text{ J s}

  2. For v=2.5×107 m s−1v = 2.5 \times 10^{7} \text{ m s}^{-1}, λ=2.9×10−11 m\lambda = 2.9 \times 10^{-11} \text{ m}: h=(2.9×10−11)×(9.11×10−31)×(2.5×107)≈6.59×10−48 J sh = (2.9 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (2.5 \times 10^{7}) \approx 6.59 \times 10^{-48} \text{ J s}

  3. For v=3.5×107 m s−1v = 3.5 \times 10^{7} \text{ m s}^{-1}, λ=2.1×10−11 m\lambda = 2.1 \times 10^{-11} \text{ m}: h=(2.1×10−11)×(9.11×10−31)×(3.5×107)≈6.61×10−48 J sh = (2.1 \times 10^{-11}) \times (9.11 \times 10^{-31}) \times (3.5 \times 10^{7}) \approx 6.61 \times 10^{-48} \text{ J s}

From these calculations, we can average the values of hh to find an approximate value.

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 fluorescent screen is indicative of an interference pattern, typically observed in wave phenomena. When the electron beam passes through the thin graphite target, it behaves similarly to light passing through a diffraction grating, creating regions of constructive and destructive interference.

The alternating bright and dark spots visible on the screen demonstrate that the electrons exhibit wave-like behavior, reinforcing the idea that they are not merely behaving as particles. If electrons were particles, we would expect to see a uniform impact distribution rather than the observed distinct pattern of light and dark areas.

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 atoms in the screen. This interaction results in the excitation of atoms, which subsequently emit photons when returning to their ground state. The fact that these collisions occur at definite points on the screen confirms that the electrons can be treated as discrete particles. Each electron impact leads to a localized burst of light, indicating the particle nature of the electrons in this interaction, separate from their wave-like behavior seen in the diffraction pattern.

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