Photo AI
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 that releases electrons from the filament is known as thermionic emission. When a current passes through the filament, it heats the metal due to its electrical resistance. As the temperature increases, the thermal energy supplied to the electrons becomes sufficient to overcome the work function of the metal, allowing them to escape from the surface. This results in the emission of electrons into the vacuum space of the apparatus.
Step 2
Show that the specific charge of the electron is given by the expression \( \frac{2V}{B^2} \).
Answer
The specific charge ( \frac{e}{m} ) can be derived from the forces acting on the electron in a magnetic field. The centripetal force needed to keep the electron in circular motion is due to the magnetic force:
- The magnetic force is given by ( F = eBv ), where ( e ) is the charge of the electron, ( B ) is the magnetic flux density, and ( v ) is the velocity of the electron.
- The centripetal force is given by ( F = \frac{mv^2}{r} ), where ( m ) is the mass, and ( r ) is the radius of the circular path.
Equating the two forces:
[ eBv = \frac{mv^2}{r} ]
Rearranging this gives:
[ e = \frac{mv}{rB} ]
For kinetic energy, the potential difference ( V ) has accelerated the electrons:
[ eV = \frac{1}{2}mv^2 \implies v^2 = \frac{2eV}{m} ]
Substituting for ( v ) into the earlier equation:
[ e = \frac{m(\frac{2eV}{m})}{rB} \implies e = \frac{2eVr}{B} ]
Rearranging gives:
[ \frac{e}{m} = \frac{2V}{B^2} ]
Step 3
Using data from Table 1, calculate a value for the specific charge of the electron.
Answer
Using the given values from Table 1:
- ( V = 320~V )
- ( B = 1.5~mT = 1.5 \times 10^{-3}~T )
Substituting these values into the equation ( \frac{e}{m} = \frac{2V}{B^2} ):
[ \frac{e}{m} = \frac{2 \times 320}{(1.5 \times 10^{-3})^2} ]
Calculating this:
[ \frac{e}{m} = \frac{640}{(2.25 \times 10^{-6})} = 2.844 \times 10^8~C/kg ]
Rounding to 3 significant figures, the specific charge of the electron is approximately ( 2.84 \times 10^8~C/kg ).
Step 4
Describe the results.
Answer
The result shows that the specific charge of the electron is significantly larger than that of the hydrogen ion, illustrating that electrons are among the lightest charged particles. This information is fundamental in particle physics, contributing to our understanding of atomic structure and the properties of matter.