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17 (a) In 1897 J J Thomson demonstrated that electrons are small negative particles - Edexcel - A-Level Physics - Question 17 - 2023 - Paper 4

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17 (a) In 1897 J J Thomson demonstrated that electrons are small negative particles. The diagram shows the apparatus used by Thomson. A potential difference V was a... show full transcript

Worked Solution & Example Answer:17 (a) In 1897 J J Thomson demonstrated that electrons are small negative particles - Edexcel - A-Level Physics - Question 17 - 2023 - Paper 4

Step 1

i) Show that the speed v of the electrons is given by v = \frac{V}{Bd}.

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Answer

To derive the expression for the speed v of the electrons, we can begin by analyzing the forces acting on the electrons. The electric field E between the plates is given by:

E=VdE = \frac{V}{d}

where V is the potential difference and d is the distance between the plates.

The force F on an electron due to this electric field can be calculated using:

F=eE=eVdF = eE = e \frac{V}{d}

where e is the charge of the electron.

In addition, the magnetic force on the electron moving with speed v in a magnetic field B is:

Fm=evF_m = ev

For the electrons to move in a straight line, these two forces must balance each other:

eVd=eve \frac{V}{d} = ev

Cancelling e from both sides, we find:

Vd=v\frac{V}{d} = v

Thus, we can rearrange this equation to show:

v=VBdv = \frac{V}{Bd}

Step 2

ii) Show that v was about 3 × 10^7 ms^-1.

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Answer

Given:

  • d = 1.5 cm = 0.015 m
  • B = 5.5 × 10^-5 T
  • V = 231 V

We can now calculate the speed v using the derived formula:

v=VBdv = \frac{V}{Bd}

Substituting the values:

v=231 V(5.5×105 T)(0.015 m)v = \frac{231 \text{ V}}{(5.5 \times 10^{-5} \text{ T})(0.015 \text{ m})}

Calculating the denominator:

Bd=5.5×105×0.015=8.25×107Bd = 5.5 \times 10^{-5} \times 0.015 = 8.25 \times 10^{-7}

Now calculating v:

v=2318.25×1072.8×108 ms1v = \frac{231}{8.25 \times 10^{-7}}\approx 2.8 \times 10^{8} \text{ ms}^{-1}

Therefore, we find that v is approximately equal to 3 × 10^7 ms^-1.

Step 3

iii) Deduce whether the charge per unit mass of an electron calculated using data from this experiment is consistent with the accepted value.

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Answer

Using the radius of the circular path of the electron:

  • r = 39 cm = 0.39 m

The centripetal force required for the circular motion is provided by the magnetic force:

mv2r=evB\frac{mv^2}{r} = evB

From this, we can express the charge-to-mass ratio (e/m) as:

em=vrB\frac{e}{m} = \frac{v}{rB}

Substituting our known values: v3×107 ms1, r=0.39 m, B=5.5×105 Tv \approx 3 \times 10^{7}\text{ ms}^{-1}, \ r = 0.39 \text{ m}, \ B = 5.5 \times 10^{-5} \text{ T}

Now calculating:

em=3×1070.39×5.5×1052.0×1011C/kg\frac{e}{m} = \frac{3 \times 10^{7}}{0.39 \times 5.5 \times 10^{-5}}\approx 2.0 \times 10^{11} C/kg

The accepted value is 1.8 × 10^{11} C/kg, which is consistent with our calculation.

Step 4

Explain how this pattern changed scientists' understanding about the nature of electrons.

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Answer

J.J. Thomson's experiments demonstrated that electrons are not just particles, but exhibit wave-like behavior as well.

The observed patterns from the electron beams interacting with thin metal films indicated interference patterns, similar to those produced by light waves. This led to the understanding that electrons can behave as waves, a concept pivotal to quantum mechanics, establishing the dual nature of electrons.

Prior to this, electrons were primarily viewed as particles; however, this experiment prompted scientists to reconsider electron behavior, contributing to the development of quantum theory and the modern understanding of atomic and subatomic structures.

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