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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
Question 1
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.
Answer
The process of electron emission at the filament is known as thermionic emission. In this process, the filament (or metal) is heated due to the current passing through it. As the temperature increases, electrons in the filament gain sufficient energy to overcome the work function of the material, allowing them to escape from the surface. This leads to the release of free electrons, which can then be accelerated towards the anode.
Step 2
Show that the specific charge of the electron is given by the expression \( \frac{2V}{B^{2}} \).
Answer
To derive the expression for the specific charge of the electron, we start with the relevant formula [ e = \frac{mv^2}{r} \text{ and } F = eB ]
Where:
- ( e ) is the charge of the electron,
- ( m ) is the mass,
- ( v ) is the velocity, ( r ) is the radius of circular motion, and ( B ) is the magnetic flux density.
Also, the kinetic energy acquired by the electrons can be expressed as
[ KE = \frac{1}{2}mv^2 = eV ]
Where ( V ) is the potential difference.
By equating the two expressions for force and rearranging, [ v = \sqrt{\frac{2eV}{m}} ]
Substituting this into the circular motion formula produces
[ eB = \frac{m(\sqrt{\frac{2eV}{m}})^2}{r} ]
[ eB = \frac{2eV}{r} ]
Finally, we can rearrange to show that
[ \frac{e}{m} = \frac{2V}{B^2} ]
Thus verifying the required expression.
Step 3
Using data from Table 1, calculate a value for the specific charge of the electron.
Answer
Given the values from Table 1:
- Potential difference ( V = 320 , V )
- Magnetic flux density ( B = 1.5 , mT = 1.5 \times 10^{-3} , T )
Using the expression derived,
[ \frac{e}{m} = \frac{2V}{B^2} ]
Substituting in the values:
[ \frac{e}{m} = \frac{2 \times 320}{(1.5 \times 10^{-3})^2} = \frac{640}{2.25 \times 10^{-6}} = 2.84 \times 10^{8} , C/kg ]
Thus, the specific charge of the electron is approximately ( 2.84 \times 10^{8} , C/kg ), presented to three significant figures.
Step 4
Describe the results.
Answer
The specific charge of the cathode rays, which are particles emitted from the filament, was found to be significantly higher than that of the hydrogen ion. This indicates that the particles constituting the cathode rays, which we understand as electrons, have a much greater charge-to-mass ratio compared to protons in a hydrogen ion. This crucial result demonstrated that electrons are much lighter than ions, confirming their fundamental role in atomic structure.