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Cathode rays and the electron questions
In the 1890s a glowing tube posed a question nobody could answer: what streams from a negative electrode through near-empty space? Measuring one ratio settled it, broke the atom's reputation as unbreakable, and handed physics its first fundamental particle.
18 original questions · 58 marks · the cathode rays and the electron notes · Turning points
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Describe how cathode rays are produced in a discharge tube.
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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).State what is meant by thermionic emission, and why the electron gun that uses it must operate in a vacuum.
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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).Define the specific charge of a particle, and state its SI unit.
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The charge of the particle divided by its mass, e/m for the electron (1). Unit: C kg−1 (1).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.
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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).An electron is accelerated from rest through a pd of 620 V. Calculate the kinetic energy it gains. e = 1.60 × 10−19 C.
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Kinetic energy gained = work done by the pd = eV (1) = 1.60 × 10−19 × 620 = 9.9 × 10−17 J (1).Using e/m = 1.76 × 1011 C kg−1, calculate the speed of electrons accelerated from rest through 3.0 kV.
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eV = ½mv2, so v = √(2(e/m)V) (1)
v = √(2 × 1.76 × 1011 × 3000) (1)
v = 3.25 × 107 m s−1 (1)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.
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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)The beam of the previous question was accelerated through 2.23 kV before entering the fields. Determine the specific charge of its particles.
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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)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.
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.
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.
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.
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.
Explain why Thomson's measurement of e/m was taken as evidence that atoms have internal structure.
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.
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.
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.
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.
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