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Figure 1 shows apparatus which can be used to determine the specific charge of an electron - AQA - A-Level Physics - Question 1 - 2018 - Paper 7

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Figure 1 shows apparatus which can be used to determine the specific charge of an electron. Electrons are emitted from the filament and accelerated by a potential d... show full transcript

Worked Solution & Example Answer:Figure 1 shows apparatus which can be used to determine the specific charge of an electron - AQA - A-Level Physics - Question 1 - 2018 - Paper 7

Step 1

Describe the process that releases the electrons emitted at the filament.

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Answer

The process that releases electrons emitted at the filament is known as thermionic emission. When an electric current passes through the filament, it heats up due to resistance. This increase in temperature supplies enough thermal energy to the electrons within the metal, allowing them to overcome the work function, which is the minimum energy required to remove an electron from the metal surface. As a result, electrons are emitted into the vacuum space of the glass tube.

Step 2

Show that the specific charge of the electron is given by the expression \( \frac{2V}{B^2} \).

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Answer

To derive the expression for the specific charge of the electron, we start from the basic principles of motion in a magnetic field. The force acting on a charged particle moving in a magnetic field is given by:

F=BqvF = Bqv

Where:

  • ( F ) is the magnetic force,
  • ( B ) is the magnetic flux density,
  • ( q ) is the charge of the particle,
  • ( v ) is the velocity of the particle.

When the electrons are accelerated through a potential difference ( V ), they gain kinetic energy given by:

KE=qV=12mv2KE = qV = \frac{1}{2} mv^2

By equating the magnetic force to the centripetal force required to keep the electron in circular motion:

mv2r=Bqvm \frac{v^2}{r} = Bqv

This leads to:

mv=Brqmv = Brq

We can rearrange this to find the velocity ( v ):

v=Brqmv = \frac{Brq}{m}

Substituting ( v ) back into the kinetic energy equation:

qV=12m(Brqm)2qV = \frac{1}{2} m \left( \frac{Brq}{m} \right)^2

Simplifying gives:

2V=B2q2m2V = \frac{B^2q^2}{m}

From this, we rearrange to express the specific charge ( \frac{q}{m} ):

qm=2VB2\frac{q}{m} = \frac{2V}{B^2}

This confirms that the specific charge of the electron is given by the expression ( \frac{2V}{B^2} ).

Step 3

Using data from Table 1, calculate a value for the specific charge of the electron.

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Answer

To calculate the specific charge of the electron, we will use the values from Table 1:

  • Potential difference, ( V = 320 V )
  • Flux density, ( B = 1.5 mT = 1.5 \times 10^{-3} T )

Now substituting these values into the equation:\n qm=2×320(1.5×103)2\frac{q}{m} = \frac{2 \times 320}{(1.5 \times 10^{-3})^2}

Calculating:\n- First calculate ( (1.5 \times 10^{-3})^2 = 2.25 \times 10^{-6} )\n- Then, calculate ( 2 \times 320 = 640 )\n So,

qm=6402.25×106=2.844×108C/kg\frac{q}{m} = \frac{640}{2.25 \times 10^{-6}} = 2.844 \times 10^8 \, C/kg

Rounding to two significant figures, the specific charge of the electron is approximately ( 2.8 \times 10^8 , C/kg ).

Step 4

Discuss the results.

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Answer

At the time of Thomson's measurements, the specific charge of particles in cathode rays was compared to that of the hydrogen ion. The results obtained indicate that the specific charge of the electron is significantly larger than that of the hydrogen ion. This points to the conclusion that electrons are much lighter compared to hydrogen ions, which consist of protons or hydrogen atoms.

This information greatly contributed to our understanding of atomic structure and the role of electrons in the atom, as the electron's specific charge allowed scientists to deduce its relative mass and charge properties compared to other known particles.

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