Motion of Charged Particles (AQA A Level Physics): Revision Note

A single proton travelling with a constant horizontal velocity enters a uniform electric field between two parallel charged plates.

The diagram shows the path taken by the proton.

Draw the path taken by a boron nucleus that enters the electric field at the same point and with the same velocity as the proton.

Atomic number of boron = 5

Mass number of boron = 11

[2]

Answer:

Step 1: Compare the charge of the boron nucleus to the proton

• Boron has 5 protons, meaning it has a charge 5 × greater than the proton

• The force on boron will therefore be 5 × greater than on the proton

• This is because electric force F is proportional to the charge Q

$F=EQ$

Step 2: Compare the mass of the boron nucleus to the proton

• The boron nucleus has a mass of 11 nucleons meaning its mass is 11 × greater than the proton

• The boron nucleus will therefore be less deflected than the proton

Step 3: Draw the trajectory of the boron nucleus

• Since the mass comparison is much greater than the charge comparison, the boron nucleus will be much less deflected than the proton [1 mark]

• The nucleus is positively charged since the neutrons in the nucleus have no charge

  • Therefore, the shape of the path will be the same as the proton

[1 mark]

Remember less deflection (e.g. from a higher mass particle) means the particle hits the plate later and the path has a smaller curve. Think of it as the particle being heavier so it is harder to steer it towards the plate.