Modern Physics — Photons, Atoms, Nuclei and Semiconductor Devices
λ = h/p. Three: an atom's energies are discrete, so a spectrum is a set of lines and not a smear. Four: mass is a form of energy, so a nucleus weighs less than its parts and the difference is what holds it together. Semiconductor devices then close the paper with the diode and its uses.Atoms and Nuclei
For hydrogen, E_n = −13.6/n² eV, so the ground state is −13.6 eV and the ionisation energy is 13.6 eV. A transition from n₂ to n₁ emits a photon of energy 13.6(1/n₁² − 1/n₂²) eV, and the series are named by where they land: Lyman on n = 1 (ultraviolet), Balmer on n = 2 (visible), Paschen on n = 3 (infrared). Radius goes as n² and speed as 1/n, which is why the electron in a higher orbit is farther out and slower. In the nucleus, the mass defect Δm is the shortfall between the sum of the free nucleons' masses and the nucleus's own, and E = Δmc² is the binding energy — with 1 u equivalent to 931.5 MeV.
| Idea | What to know |
|---|---|
| Binding energy per nucleon | Peaks near iron-56 at about 8.8 MeV, which is why both fusion of light nuclei and fission of heavy ones release energy — each moves towards the peak. |
| Alpha decay | Z falls by 2 and A by 4 — a helium nucleus leaves. |
| Beta-minus decay | Z rises by 1 and A is unchanged: a neutron becomes a proton, with an electron and an antineutrino emitted. |
| Gamma emission | Neither Z nor A changes; the nucleus simply drops to a lower energy state. |
| Half-life | After n half-lives the fraction left is (1/2)ⁿ, and the half-life of a sample never depends on how much of it there is. |
| Nuclear density | Radius goes as A^(1/3), so volume goes as A and the density is the SAME for every nucleus. |
Semiconductor Devices, as the Current Book Has Them
Doping a pure semiconductor with a pentavalent atom gives an n-type with electrons as majority carriers; a trivalent atom gives a p-type with holes as majority. Both remain electrically neutral — doping adds carriers, not charge, and that is a favourite one-mark question. A p-n junction has a depletion region and a barrier potential (about 0.7 V for silicon, 0.3 V for germanium); it conducts on forward bias and blocks on reverse bias, which makes it a rectifier: a half-wave rectifier uses one diode and passes one half of each cycle, a full-wave rectifier uses two and passes both, so its output frequency is twice the input's.
- Zener diode works in reverse breakdown at a sharply defined voltage, which makes it a voltage regulator — the one diode deliberately used in reverse.
- Photodiode is also used in reverse bias, because the reverse current is what responds to light. LED is forward-biased and emits when electrons and holes recombine across the junction.
- Solar cell needs no external bias at all: it generates an emf from incident light.
- What the rationalised book no longer carries: the junction transistor and logic gates, and Communication Systems as a chapter. Check the year's NTA syllabus before deciding to skip them.
Key takeaways
- hν = φ + K_max: below the threshold frequency nothing is emitted however bright the light, and above it the electrons' energy depends on frequency alone while intensity fixes only their number. Emission is instantaneous.
- λ = h/p, so at equal kinetic energy a heavier particle has the shorter wavelength. For hydrogen E_n = −13.6/n² eV, with Lyman landing on n = 1 in the ultraviolet and Balmer on n = 2 in the visible.
- Binding energy per nucleon peaks near iron-56 at about 8.8 MeV, which is why both fusion and fission release energy. Nuclear radius goes as A^(1/3), so nuclear density is the same for every nucleus, and after n half-lives (1/2)ⁿ of a sample is left.
- Doping adds carriers, not charge — both n-type and p-type stay neutral. Forward bias for the rectifier diode and the LED, reverse for the Zener and the photodiode, and no bias at all for the solar cell.
Practice questions (8)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In a photoelectric experiment the intensity of the incident light is doubled at unchanged frequency. Which quantity doubles?
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Answer: B — The number of electrons emitted per second
Intensity is photons per second, so doubling it doubles the electrons ejected per second and therefore the photocurrent, while each photon still carries hν and gives each electron the same maximum energy. Maximum kinetic energy and stopping potential are two names for that same energy — eV₀ = K_max — so both are set by frequency alone and both stay put; offering them separately is how this item catches a candidate who has the rule the wrong way round. And the threshold frequency is a property of the metal's work function, which no amount of light can change.The ionisation energy of a hydrogen atom in its ground state is:
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Answer: B — 13.6 eV
The ground-state energy is E₁ = −13.6 eV, so removing the electron to n = ∞, where the energy is zero, costs 13.6 eV. 3.4 eV is |E₂|, the energy of the first excited state, and it is the ionisation energy from n = 2 rather than from the ground state. 10.2 eV is the n = 1 to n = 2 excitation energy, 13.6 − 3.4, which is the most-chosen wrong answer because it is the other famous hydrogen number. And 1.51 eV is |E₃|.A radioactive sample has a half-life of 10 days. The fraction remaining after 30 days is:
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Answer: B — 1/8
Thirty days is three half-lives, and the fraction left after n half-lives is (1/2)ⁿ = 1/8. Decay is exponential rather than linear, which is why 1/3 — dividing the time by the half-life and treating the result as a fraction destroyed — is wrong, and it is the commonest error here. 1/6 corresponds to no rule at all. And zero would require decay to finish, which an exponential never does: after every further half-life half of whatever remains is still there.In beta-minus decay, the atomic number Z and the mass number A of the nucleus change as:
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Answer: B — Z rises by 1, A is unchanged
A neutron inside the nucleus becomes a proton, emitting an electron and an antineutrino, so the proton count rises by one while the total number of nucleons is untouched. "Z falls by 1" describes beta-PLUS decay or electron capture and is chosen by candidates who reason that emitting a negative charge should lower Z — but the charge leaves the nucleus while a proton is created inside it. Z down 2 and A down 4 is alpha decay. And both unchanged is gamma emission, where the nucleus only sheds energy.A semiconductor doped with a pentavalent impurity is:
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Answer: B — n-type, and electrically neutral overall
A pentavalent atom brings five valence electrons where four are needed for bonding, so the fifth is nearly free: the material is n-type with electrons as majority carriers. But the donor atom also brings a proton for every electron, so the crystal stays neutral — doping supplies mobile carriers, not net charge. "n-type and negatively charged" gets the type right and the charge wrong, and it is the most-chosen wrong option because "n" suggests negative. The two p-type options invert the valency rule: it is a TRIVALENT impurity that leaves holes and gives p-type.Which device is normally operated in REVERSE bias?
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Answer: B — A Zener diode used as a voltage regulator
A Zener is designed to work in reverse breakdown, where the voltage across it stays almost constant over a wide range of current — which is exactly what a regulator needs. An LED must be forward-biased, since it emits when electrons and holes recombine across the junction. A rectifier diode conducts in forward bias, which is the half-cycle the question names. And a solar cell is the outlier: it uses no external bias at all, generating its own emf from light, so "reverse" is wrong for a different reason than the others.A proton and an electron have the same kinetic energy. Comparing their de Broglie wavelengths:
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Answer: B — The proton’s is much shorter, because p = √(2mK) is larger for the heavier particle
λ = h/p and p = √(2mK), so at equal K the wavelength goes as 1/√m: the proton, about 1836 times heavier, has a wavelength roughly 43 times shorter. Equal energies giving equal wavelengths ignores the mass in the momentum and is the intuitive wrong answer. "Longer because it carries more momentum" states the momentum comparison correctly and then inverts the relation, since λ is inversely proportional to p — a good example of a right fact used backwards. And the last option denies the de Broglie hypothesis itself, which is the premise of the whole unit.Both the fusion of light nuclei and the fission of heavy nuclei release energy because:
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Answer: B — Binding energy per nucleon peaks near iron, so both move their products towards that peak
The binding-energy-per-nucleon curve rises steeply to about 8.8 MeV near iron-56 and falls slowly after it. Light nuclei fusing move up the rising side and heavy nuclei splitting move back towards the peak from the falling side, and in both directions the products are more tightly bound, so the mass defect appears as released energy. Exact mass conservation is the opposite of what happens — some mass becomes energy through E = Δmc². Nucleon number is conserved rather than increased. And radioactivity is a separate property that explains neither process.