Question 1:
(a) What are thermionic emission and cathode rays?
(b) State the characteristics of cathode rays.
Answer:
(a) Thermionic emission is the emission of electrons from a heated metallic surface. The cathode rays are high speed electron beams in a vacuum tube.
(b) Cathode rays are negatively charged, possess momentum and kinetic energy, move in straight lines and can be deflected by electric fields and magnetic fields.
(a) What are thermionic emission and cathode rays?
(b) State the characteristics of cathode rays.
Answer:
(a) Thermionic emission is the emission of electrons from a heated metallic surface. The cathode rays are high speed electron beams in a vacuum tube.
(b) Cathode rays are negatively charged, possess momentum and kinetic energy, move in straight lines and can be deflected by electric fields and magnetic fields.
Question 2:
(a) State the function of the components of a cathode ray tube below:
(i) heating filament
(iii) anode
(ii) cathode
(iv) fluorescent screen
(b) Why must a cathode ray tube be in a state of vacuum?
Answer:
(a)(i) Heats the cathode to a high temperature
(a)(ii) Emits electrons (thermionic emission)
(a)(iii) Accelerates the electron beam until it reaches a high velocity
(a)(iv) Produces light spots when high-velocity electron beams hit the fluorescent screen
(b) So that the electrons do not collide with the air molecules.
(a) State the function of the components of a cathode ray tube below:
(i) heating filament
(iii) anode
(ii) cathode
(iv) fluorescent screen
(b) Why must a cathode ray tube be in a state of vacuum?
Answer:
(a)(i) Heats the cathode to a high temperature
(a)(ii) Emits electrons (thermionic emission)
(a)(iii) Accelerates the electron beam until it reaches a high velocity
(a)(iv) Produces light spots when high-velocity electron beams hit the fluorescent screen
(b) So that the electrons do not collide with the air molecules.
Question 3:
When an electron beam moves from the cathode to the anode in a vacuum tube, state:
(a) the type of motion of the electron beam
(b) the transformation of energy
(c) the relationship between the voltage of E.H.T. power supply and the velocity of the electron.
Answer:
(a) Uniform acceleration
(b) Electrical potential energy is converted into kinetic energy of electron
(c)
$$ \begin{aligned} e V & =\frac{1}{2} m v_{\max }^2 \\ \text { that is } e & =\text { charge of an electron }\left(1.6 \times 10^{-19} \mathrm{C}\right) \\ V & =\text { potential difference between cathode and } \\ & \text { anode } \\ m & =\text { mass of electron }\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \\ V_{\max } & =\text { maximum velocity of electron } \end{aligned} $$
When an electron beam moves from the cathode to the anode in a vacuum tube, state:
(a) the type of motion of the electron beam
(b) the transformation of energy
(c) the relationship between the voltage of E.H.T. power supply and the velocity of the electron.
Answer:
(a) Uniform acceleration
(b) Electrical potential energy is converted into kinetic energy of electron
(c)
$$ \begin{aligned} e V & =\frac{1}{2} m v_{\max }^2 \\ \text { that is } e & =\text { charge of an electron }\left(1.6 \times 10^{-19} \mathrm{C}\right) \\ V & =\text { potential difference between cathode and } \\ & \text { anode } \\ m & =\text { mass of electron }\left(9.1 \times 10^{-31} \mathrm{~kg}\right) \\ V_{\max } & =\text { maximum velocity of electron } \end{aligned} $$
Question 4:
When an E.H.T. with power of 800 V is connected across the cathode and the anode, what is the velocity of the electron? What is the effect on the velocity of the electron if the voltage is increased by four times?
[Charge of an electron, e = 1.6 × 10–19 C, mass of an electron, m = 9.11 × 10–31 kg]
Answer:
$$ \begin{aligned} V & =800 \mathrm{~V} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m & =9.11 \times 10^{-31} \mathrm{~kg} \\ v & =\sqrt{\frac{2 e V}{m}} \\ & =\sqrt{\frac{2 \times\left(1.6 \times 10^{-19}\right) \times(800)}{9.1 \times 10^{-31}}} \\ & =1.68 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ v=\sqrt{\frac{2 e V}{m}} \Rightarrow 2 v=\sqrt{\frac{2 e(4 V)}{m}} $$
When an E.H.T. with power of 800 V is connected across the cathode and the anode, what is the velocity of the electron? What is the effect on the velocity of the electron if the voltage is increased by four times?
[Charge of an electron, e = 1.6 × 10–19 C, mass of an electron, m = 9.11 × 10–31 kg]
Answer:
$$ \begin{aligned} V & =800 \mathrm{~V} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m & =9.11 \times 10^{-31} \mathrm{~kg} \\ v & =\sqrt{\frac{2 e V}{m}} \\ & =\sqrt{\frac{2 \times\left(1.6 \times 10^{-19}\right) \times(800)}{9.1 \times 10^{-31}}} \\ & =1.68 \times 10^7 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
$$ v=\sqrt{\frac{2 e V}{m}} \Rightarrow 2 v=\sqrt{\frac{2 e(4 V)}{m}} $$
The electron velocity will be doubled if the potential difference is increased by four times.