A full lesson: a cartoon movie to start, then the glowing discharge tube, Thomson's crossed fields and e/m, Millikan's oil drop and charge quantisation, the work function and the three ways to free an electron, the electron volt, NEET shortcuts, problems to solve yourself, and 12 MCQs.
Dadi's old TV is a cathode ray tube: a hot cathode boils off electrons, a high voltage fires them at the screen, and a magnet can bend the beam because it is made of moving negative charges.
Kabir and Meera find Dadi's old television in the store room: a heavy box with a curved glass face. When Kabir switches it on, the glass glows before any picture appears. Inside is a cathode ray tube: a glass bottle with almost no air, a hot cathode at the back and a glowing screen at the front. Everything you see on such a screen is drawn by a beam of tiny charged particles hitting the glass.
A hundred and fifty years ago, nobody knew what that beam was. Physicists sent electricity through gases at very low pressure, saw the glass glow, and called the invisible cause cathode rays. Finding out what they were gave us the electron, the first particle smaller than an atom, and the idea of a work function that controls how electrons leave a metal. That is this lesson.
"Every electron in this chapter has to escape from a metal first. Heat it, pull it with a strong field, or shine light on it. Remember those three doors: thermionic, field and photoelectric emission."
"Look closely at the two ends of an old tube light. Why are they often blackened, and why do they glow orange for a moment when you switch it on?"
Each end has a coiled filament coated with a material of low work function. When it is heated it gives out electrons (thermionic emission) to start the discharge. Over the years some of the coating evaporates and darkens the glass. Please do not bring magnets near a picture-tube TV you want to keep; it can permanently spoil the colours.
Air at normal pressure does not conduct. Seal two electrodes in a glass tube, apply a high voltage and pump the air out. Slide the pressure down from 760 mm of mercury and watch. At about 0.001 mm of Hg the gas stops glowing, and the glass wall opposite the cathode glows (yellowish-green for soda glass). Something leaves the cathode and hits the glass. Then bring a magnet.
The glass facing the cathode fluoresces, and an object in the way casts a sharp shadow: the rays travel in straight lines from the cathode.
An electric field pulls them towards the positive plate and a magnet bends them. So they are streams of fast negatively charged particles, not light.
Change the cathode metal or the gas: the particles behave the same way. They must be a part of all matter.
"NCERT gives you three numbers to keep: pressure about ten to the minus three millimetre of mercury, glow yellowish-green on soda glass, and Crookes, 1870 and 1879."
J. J. Thomson fired the beam between two charged plates (field E, pushes the electrons towards +) and put a magnetic field B at right angles to it, pushing the other way. Change E, B and the gun voltage V. When the two forces balance, the spot comes back to the centre of the screen.
With no deflection, the electric push equals the magnetic push. This measures the speed. Thomson found speeds of about 0.1 c to 0.2 c.
The electron gained its kinetic energy by falling through the gun voltage V.
The same value for every cathode metal and every gas. In 1897 Thomson named the particle the electron.
"Crossed fields are a velocity selector. Only particles with v equal to E by B go straight, whatever their charge or mass. NEET loves this one line."
"If we double both E and B, does the beam still go straight?"
Yes. Only the ratio E/B matters for the balance, and doubling both keeps it the same. But the speed still has to match: v = E/B is unchanged, so the same beam passes straight.
A tiny charged oil drop (mass 6.54 × 10⁻¹⁵ kg) floats between two plates 1 cm apart. Gravity pulls it down, the electric force qE pushes it up. Slide the plate voltage until the drop hangs still: then qE = mg and you can work out its charge. Zap it with X-rays to change its charge and measure again.
Hanging still means no net force (the air drag is zero when the drop is not moving).
Here m g d = 6.54 × 10⁻¹⁵ × 9.8 × 0.01 = 6.41 × 10⁻¹⁶ J, so q = 6.41 × 10⁻¹⁶ / V.
Every measured charge is a whole number times e = 1.602 × 10⁻¹⁹ C. Charge is quantised.
"Look at the dots on the charge ruler. They never land between the marks. Nature does not sell half an electron."
A metal is full of free electrons, yet they do not fly out. If one leaves, the metal is left positive and pulls it back. The electrons sit in an energy well: the minimum energy needed to lift one out of the surface is the work function φ₀. Choose a metal and a way to supply the energy.
| Emission | Energy comes from | Where you meet it |
|---|---|---|
| Thermionic | heat: thermal energy of the electrons | filament of a picture tube, X-ray tube, tube-light ends |
| Field | a very strong field, about 10⁸ V/m | sharp tips in electron microscopes, sparks |
| Photoelectric | light of high enough frequency | photocells, light meters (next lessons) |
"Work function belongs to the metal and to its surface, not to the light or the heat you supply. Caesium has the smallest one in the NCERT table, 2.14 eV, platinum one of the largest, 5.65 eV."
"Why do picture tubes coat their cathode with oxides of barium and strontium?"
Those coatings have a low work function, so a moderately hot cathode gives out plenty of electrons. A lower φ₀ means less heating is needed for the same emission.
One electron volt (1 eV) is the energy an electron gains when it is accelerated through a potential difference of 1 volt: 1 eV = 1.602 × 10⁻¹⁹ J. Pick a particle and a voltage, release it from rest, and read its energy and speed.
Electron or proton through V volt: K = V eV. Alpha (charge 2e): K = 2V eV. The mass does not matter for the energy.
For an electron, v ≈ 5.93 × 10⁵ √V m/s. 100 V gives 5.93 × 10⁶ m/s. Much heavier particles go much slower for the same energy.
In joule these are about 10⁻¹⁹: awkward. Atomic and photon energies are easy numbers in eV. Convert before mixing units.
eV → J: multiply. J → eV: divide. 1 keV = 10³ eV, 1 MeV = 10⁶ eV.
Undeflected beam in crossed fields. Independent of charge and mass. E = 3 × 10⁴ V/m, B = 10⁻³ T → v = 3 × 10⁷ m/s.
From balance plus gun: e/m = E²/(2VB²). For a proton e/m is 1836 times smaller.
Divide by 1.6 × 10⁻¹⁹. 2.4 × 10⁻¹⁹ C gives 1.5: impossible charge.
Non-relativistic, good up to a few kV. 400 V → 20 × 5.93 × 10⁵ = 1.19 × 10⁷ m/s.
Electron and alpha through the same V: K ratio 1 : 2. Proton and alpha: K 1 : 2, v ratio √2 : 1.
"Units are where students lose marks in this chapter. Work functions, photon energies and kinetic energies: put them all in eV first, and only change to joule when a mass or a speed appears."
Cover the solution, write the formula, and solve on paper. Open the hint only if you are stuck for more than a minute. Take e = 1.6 × 10⁻¹⁹ C where a rounded answer is enough.
A metal sphere carries a charge of −1.0 μC. How many extra electrons does it have?
q = n e, so n = q / e.
In Thomson's apparatus, an electron beam goes straight through crossed fields E = 4.5 × 10⁴ V/m and B = 1.5 × 10⁻³ T. Find the speed of the electrons. What fraction of c is it?
Balanced: eE = evB.
An electron starts from rest and is accelerated through 500 V. Find its kinetic energy in eV and in joule, and its speed (m = 9.11 × 10⁻³¹ kg).
K = eV, then ½ m v² = K.
An oil drop of mass 9.8 × 10⁻¹⁵ kg hangs still between two plates 1.0 cm apart when the voltage across them is 2000 V. How many excess electrons does it carry? (g = 9.8 m/s²)
qE = mg with E = V/d.
An alpha particle (charge +2e) is accelerated from rest through 1 kV. Find its kinetic energy in eV and in joule. Compare with an electron through the same voltage.
K = qV with q = 2e.
Take e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg. Write the formula first, then click.
Score: 0 / 12
| If the question says… | Write this first | The trap |
|---|---|---|
| "undeflected in crossed E and B" | v = E / B | Writing B/E, or bringing in the charge or the mass |
| "accelerated through V volt" | K = qV (in eV: q in units of e × V) | Forgetting the 2e of an alpha particle |
| "speed of an electron after V volt" | v = √(2eV/m) ≈ 5.93 × 10⁵ √V | Using 2 eV/m without the square root |
| "which charge is not possible" | n = q/e must be a whole number | Accepting 1.5e or 2.5e |
| "oil drop held stationary" | qV/d = mg | Using V in place of the field E = V/d |
| "work function depends on" | metal and its surface condition | Choosing the intensity or frequency of light |
| "energy in joule" from eV | × 1.602 × 10⁻¹⁹ | Dividing instead of multiplying |
| "cathode rays are" | fast electrons (negative particles) | Calling them electromagnetic waves |
Lesson 2 shines light on a metal plate: the photoelectric effect. The work function you met here decides the threshold frequency, and the electron volt measures the energy of the photoelectrons.
Values used: e = 1.602 × 10⁻¹⁹ C, m(electron) = 9.11 × 10⁻³¹ kg, e/m = 1.76 × 10¹¹ C/kg, m(proton) = 1.673 × 10⁻²⁷ kg, g = 9.8 m/s², work functions from NCERT Table 11.1 plus tungsten ≈ 4.5 eV (a standard value).
Follow the four steps: understand it, remember it, try it, then prove it with past papers.