Edexcel iGCSE Physics (4PH1) 6.1 Units Exam Style Question Paper 2B - New Syllabus
Question
The diagram shows the inside of an oscilloscope. Electrons leave the electron supply and accelerate towards the screen. The stream of electrons hits the screen.
(a) The grid is connected to the positive terminal of a high voltage power supply.
(i) Explain, in terms of the movement of particles, how the grid has become positively charged.
(ii) State the formula linking kinetic energy, mass and speed.
(iii) Calculate the speed of an electron when it has \(1.3\times10^{-15}\,\mathrm{J}\) of kinetic energy. The mass of an electron is \(9.1\times10^{-31}\,\mathrm{kg}\).
(b) The stream of electrons spreads out as it travels towards the screen. Explain why the electrons in the stream move apart from each other.
(c) The oscilloscope can cause the electrons to move in a circle. This produces a circular pattern on the screen.
(i) Use the scale to determine the radius of the circle.
(ii) The time taken for the electrons to complete one orbit of the circle is \(24\,\mathrm{ms}\). Calculate the orbital speed of the electrons. Use the formula
\(\displaystyle \mathrm{orbital\ speed}=\frac{2\pi\times\mathrm{orbital\ radius}}{\mathrm{time\ period}}\)
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 4.14: Kinetic Energy — parts (a)(ii)–(iii)
• 2.25P: Forces Between Electric Charges — part (b)
• 6.11P: Force on a Charged Particle in a Magnetic Field — part (c)
▶️ Answer/Explanation
(a)(i) Positive charging [2 marks]
The grid becomes positively charged because it loses electrons.
Electrons are transferred away from the grid, leaving it with more positive charge than negative charge.
(a)(ii) Kinetic energy formula [1 mark]
The formula linking kinetic energy, mass and speed is:
\(\boxed{E_\mathrm{k}=\dfrac{1}{2}mv^2}\)
(a)(iii) Speed of the electron [3 marks]
Use:
\(E_\mathrm{k}=\dfrac{1}{2}mv^2\)
Substitute the values:
\(1.3\times10^{-15}=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2\)
Rearranging:
\(\displaystyle v^2=\frac{2(1.3\times10^{-15})}{9.1\times10^{-31}}\)
\(v^2\approx2.86\times10^{15}\)
Therefore:
\(v=\sqrt{2.86\times10^{15}}\)
\(v\approx5.35\times10^7\,\mathrm{m\,s^{-1}}\)
\(\boxed{v=5.3\times10^7\,\mathrm{m\,s^{-1}}}\)
(b) Repulsion between electrons [2 marks]
All the electrons in the stream have the same type of charge, so they are like charges.
Like charges repel each other. Therefore, the electrons exert repulsive forces on one another and move apart as they travel towards the screen.
(c)(i) Radius of the circle [1 mark]
From the scale on the diagram, the radius of the circular pattern is:
\(\boxed{r=2.5\,\mathrm{cm}}\)
(c)(ii) Orbital speed [3 marks]
Convert the quantities into SI units:
\(r=2.5\,\mathrm{cm}=2.5\times10^{-2}\,\mathrm{m}\)
\(T=24\,\mathrm{ms}=24\times10^{-3}\,\mathrm{s}\)
Use:
\(\displaystyle v=\frac{2\pi r}{T}\)
\(\displaystyle v=\frac{2\pi(2.5\times10^{-2})}{24\times10^{-3}}\)
\(v\approx6.54\,\mathrm{m\,s^{-1}}\)
Therefore:
\(\boxed{v=6.5\,\mathrm{m\,s^{-1}}}\)