Lesson 4: Physics: Modern Physics and Semiconductors
Modern physics asks what classical physics cannot explain: why light ejects electrons only above a threshold frequency, why atoms emit only certain colours, and why nuclei decay at a fixed rate. Each of these has a clean quantitative law, and NEET questions test whether you can apply it.
This lesson covers the photoelectric effect, Bohr's atomic model, radioactivity and semiconductors. The constants to keep at hand are Planck's constant h ≈ 6.63 × 10⁻³⁴ J·s, the electron charge e = 1.6 × 10⁻¹⁹ C, and the handy conversion hc ≈ 1240 eV·nm for photon energies.
A useful habit: convert photon energies to electron-volts early. Wavelengths in nanometres divided into 1240 give energies in eV directly, which avoids powers of ten in the middle of a calculation.
Photoelectric effect and matter waves
Einstein's photoelectric equation: the maximum kinetic energy of an ejected electron is Kmax = hf − φ, where φ is the work function of the metal. Below the threshold frequency f₀ = φ/h, no electrons are ejected however bright the light — intensity changes the number of electrons, not their maximum energy.
A photon of frequency 5 × 10¹⁴ Hz has energy E = hf = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ = 3.315 × 10⁻¹⁹ J.
Light of wavelength 620 nm has photon energy E = 1240/620 = 2 eV.
Light of photon energy 4.0 eV striking a metal of work function 2.3 eV ejects electrons with Kmax = 4.0 − 2.3 = 1.7 eV.
Light of wavelength 400 nm has photon energy 1240/400 = 3.1 eV.
De Broglie's wavelength for a particle of momentum p is λ = h/p. For an electron of mass about 9.11 × 10⁻³¹ kg moving at 1 × 10⁶ m/s, p ≈ 9.11 × 10⁻²⁵ kg·m/s, so λ ≈ 6.63 × 10⁻³⁴ / 9.11 × 10⁻²⁵ ≈ 7.3 × 10⁻¹⁰ m — comparable to atomic spacings, which is why electron diffraction works.
Bohr model and atomic spectra
In Bohr's model, the energy of level n in hydrogen is En = −13.6/n² eV. The energy of a photon emitted in a transition is the difference between levels, and its wavelength follows from E = hc/λ.
For n = 2: E₂ = −13.6/4 = −3.4 eV.
For n = 3: E₃ = −13.6/9 ≈ −1.51 eV.
A transition from n = 3 to n = 2 releases 3.4 − 1.51 ≈ 1.89 eV, in the visible range.
Spectral series are named by their lower level: Lyman ends at n = 1 (ultraviolet), Balmer at n = 2 (visible), Paschen at n = 3 (infrared). If a question asks for the series of a visible line, the answer is Balmer.
Nuclei, radioactivity and semiconductors
Radioactive decay follows N = N₀(1/2)^(t/T½), where T½ is the half-life. After one half-life half remains, after two a quarter, after three an eighth. Binding energy per nucleon peaks near iron, which is why both fusion of light nuclei and fission of heavy nuclei can release energy.
A sample of 80 g has a half-life of 10 days. After 30 days, three half-lives have passed.
Remaining mass = 80 × (1/2)³ = 80 × 1/8 = 10 g.
In semiconductors, doping with a pentavalent impurity gives an n-type semiconductor (electrons majority carriers); doping with a trivalent impurity gives p-type (holes majority carriers). A p–n junction diode conducts easily in forward bias and blocks current in reverse bias (apart from a tiny leakage current), and a Zener diode is designed to operate in reverse breakdown as a voltage regulator.
Common pitfalls
Do not confuse light intensity with photon energy in the photoelectric effect: intensity changes the number of electrons; frequency changes their maximum energy.
Do not forget that Bohr energies are negative and get less negative (higher) as n increases: E₃ (−1.51 eV) is higher than E₂ (−3.4 eV).
Further study
- NCERT Class 12 Physics, chapters on dual nature of radiation and matter, atoms, nuclei and semiconductor electronics.
📝 Quiz
1. A photon of frequency 5 × 10¹⁴ Hz has what energy (h = 6.63 × 10⁻³⁴ J·s)?
Show answer
a) E = hf = 6.63 × 10⁻³⁴ × 5 × 10¹⁴ = 3.315 × 10⁻¹⁹ J.
2. Light of 4.0 eV hits a metal of work function 2.3 eV. What is Kmax of the ejected electrons?
Show answer
b) Kmax = hf − φ = 4.0 − 2.3 = 1.7 eV.
3. An 80 g radioactive sample (half-life 10 days) is left for 30 days. How much remains?
Show answer
c) 30 days is three half-lives: 80 × (1/2)³ = 80/8 = 10 g.
4. What is the energy of the n = 2 level of hydrogen in Bohr's model?
Show answer
c) En = −13.6/n², so E₂ = −13.6/4 = −3.4 eV.