Questions on cathode rays and the electron gun

Question 1

Describe how cathode rays are produced in a discharge tube.

Accepted Answer

A few kilovolts across a low-pressure gas ionises it; positive ions strike the cathode and release electrons (1), which accelerate towards the anode, making the gas glow along their path (1).

Question 2

State what is meant by thermionic emission, and why the electron gun that uses it must operate in a vacuum.

Accepted Answer

Electrons in a heated metal gain enough kinetic energy to escape its surface (1). In air the beam would collide with gas molecules and scatter, so the gun is evacuated (1).

Question 3

Define the specific charge of a particle, and state its SI unit.

Accepted Answer

The charge of the particle divided by its mass, e/m for the electron (1). Unit: C kg⁻¹ (1).

Question 4

A bar magnet is brought close to a working discharge tube and the glowing beam shifts sideways. State two conclusions about the nature of cathode rays that follow from observations of this kind.

Accepted Answer

The rays are deflected by a magnetic field, so they are moving charged particles (1). The direction of the deflection shows that the charge is negative (1).

Question 5

An electron is accelerated from rest through a pd of 620 V. Calculate the kinetic energy it gains. e = 1.60 × 10⁻¹⁹ C.

Accepted Answer

Kinetic energy gained = work done by the pd = eV (1) = 1.60 × 10⁻¹⁹ × 620 = 9.9 × 10⁻¹⁷ J (1).

Question 6

Using e/m = 1.76 × 10¹¹ C kg⁻¹, calculate the speed of electrons accelerated from rest through 3.0 kV.

Accepted Answer

eV = ½mv², so v = √(2(e/m)V) (1) v = √(2 × 1.76 × 10¹¹ × 3000) (1) v = 3.25 × 10⁷ m s⁻¹ (1)

Question 7

An electron beam passes undeflected through crossed fields of strength E = 4.2 × 10⁴ V m⁻¹ and B = 1.5 × 10⁻³ T. Calculate the beam speed and explain why no deflection occurs.

Accepted Answer

v = E/B = 4.2 × 10⁴/(1.5 × 10⁻³) (1) v = 2.8 × 10⁷ m s⁻¹ (1) The electric force eE and magnetic force Bev are equal and opposite at exactly this speed (1)

Question 8

The beam of the previous question was accelerated through 2.23 kV before entering the fields. Determine the specific charge of its particles.

Accepted Answer

eV = ½mv² (1) Rearranging: e/m = v²/2V (1) = (2.8 × 10⁷)²/(2 × 2230) (1) e/m = 1.76 × 10¹¹ C kg⁻¹ (1)

Question 9

Calculate the specific charge of the hydrogen ion (mass 1.67 × 10⁻²⁷ kg, charge 1.60 × 10⁻¹⁹ C), and the ratio of the electron's specific charge to it.

Accepted Answer

e/m = 1.60 × 10⁻¹⁹/(1.67 × 10⁻²⁷) (1) e/m = 9.58 × 10⁷ C kg⁻¹ (1) Ratio: 1.76 × 10¹¹/(9.58 × 10⁷) ≈ 1840 (1)

Question 10

In an electron gun, electrons are accelerated from rest through 730 V. Show that the electrons leave the gun at about 1.6 × 10⁷ m s⁻¹. Go on to calculate the time the electrons take to cross a pair of deflecting plates 60 mm long. e/m = 1.76 × 10¹¹ C kg⁻¹.

Accepted Answer

v = √(2(e/m)V) (1) = √(2 × 1.76 × 10¹¹ × 730) = 1.60 × 10⁷ m s⁻¹, about 1.6 × 10⁷ m s⁻¹ (1). Time = length/speed = 0.060/(1.60 × 10⁷) (1) = 3.7 × 10⁻⁹ s (1).

Question 11

Electrons accelerated from rest through 3.2 kV enter the space between parallel plates 4.0 mm apart with a pd of 460 V across them. A magnetic field at right angles to both the beam and the electric field lets the beam pass undeflected. Calculate the magnetic flux density required. e/m = 1.76 × 10¹¹ C kg⁻¹.

Accepted Answer

v = √(2(e/m)V) = √(2 × 1.76 × 10¹¹ × 3200) = 3.36 × 10⁷ m s⁻¹ (1). E = V/d = 460/(4.0 × 10⁻³) = 1.15 × 10⁵ V m⁻¹ (1). Undeflected means Bev = eE, so B = E/v (1) = 3.4 × 10⁻³ T (1).

Question 12

In a discharge tube containing a trace of gas, a beam of positive ions is found to have a specific charge of 4.79 × 10⁷ C kg⁻¹. Each ion carries a single charge of 1.60 × 10⁻¹⁹ C. Calculate the mass of one ion and, by comparison with the hydrogen ion (mass 1.67 × 10⁻²⁷ kg), suggest what the ions could be.

Accepted Answer

m = charge ÷ specific charge (1) = 1.60 × 10⁻¹⁹/(4.79 × 10⁷) = 3.34 × 10⁻²⁷ kg (1). This is 2.0 times the hydrogen ion's mass (1), so the ions could be singly charged particles of twice its mass, for example hydrogen molecule ions H₂⁺ (or deuterium ions) (1).

Question 13

Calculate the accelerating pd that gives electrons a speed of one tenth of the speed of light, and suggest why the simple formula should be used cautiously beyond this speed.

Accepted Answer

V = v²/(2 e/m) (1)
= (3.0 × 10⁷)²/(2 × 1.76 × 10¹¹) (1)
V = 2557 V, about 2.6 kV (1)
Beyond a tenth of c, relativistic effects grow: the work done increasingly raises the electron's mass-energy rather than its speed, so eV = ½mv² over-predicts v (1)

Question 14

Explain why Thomson's measurement of e/m was taken as evidence that atoms have internal structure.

Accepted Answer

The specific charge was over 1000 times that of the hydrogen ion (about 1800 by modern values), the largest previously known (1), and identical whatever cathode metal or gas was used (1). Assuming its charge is comparable to the hydrogen ion's, the particle must be far lighter than the lightest atom, and common to all matter: a universal constituent of atoms (1).

Question 15

In a specific charge measurement, a student suggests removing the electric field and finding the speed from the magnetic deflection alone. Explain why the crossed-fields arrangement is used instead to find the speed.

Accepted Answer

With only a magnetic field, the deflection depends on both the speed and the unknown specific charge (1), so the two cannot be separated (1). Balancing the two fields gives the speed directly as v = E/B, independent of e and m, which can then be combined with the accelerating pd (1).

Question 16

A student builds a crossed-fields apparatus. The beam, accelerated from rest through 1.92 kV, passes undeflected when E = 3.90 × 10⁴ V m⁻¹ and B = 1.50 × 10⁻³ T. Deduce whether the beam consists of electrons. Specific charge of the electron = 1.76 × 10¹¹ C kg⁻¹.

Accepted Answer

v = E/B = 3.90 × 10⁴/(1.50 × 10⁻³) = 2.60 × 10⁷ m s⁻¹ (1). Specific charge = v²/2V (1) = (2.60 × 10⁷)²/(2 × 1920) = 1.76 × 10¹¹ C kg⁻¹ (1). This matches the electron's specific charge, so the beam is consistent with being electrons (1).

Question 17

Describe how the specific charge of the electron can be determined using an electron gun and crossed electric and magnetic fields. Your answer should include the quantities measured and how e/m is calculated from them.

Accepted Answer

Electrons from a heated filament are accelerated from rest through a measured pd V (1). The beam enters an electric field between parallel plates, crossed with a magnetic field at right angles to both the beam and the electric field (1). One field is adjusted until the beam is undeflected, when the electric and magnetic forces balance: eE = Bev (1). The beam speed follows as v = E/B (1). Since eV = ½mv², the specific charge is e/m = v²/2V (1). Measured quantities: the accelerating pd, E (from the plate pd and separation) and B (1).

Question 18

The pd of an electron gun is increased from 500 V to 1000 V. A student predicts that the electron speed doubles. Deduce the factor by which the speed actually increases, and the pd that would be needed to double it.

Accepted Answer

v = √(2(e/m)V), so v ∝ √V (1). Doubling the pd therefore raises the speed by a factor √2 ≈ 1.4, not 2 (1). Doubling the speed needs four times the pd: 2000 V (1).