Modern physics JEE questions: formulas, patterns and the mistakes that cost marks

Photons carry energy in packets

A photon of frequency f carries E = hf = hc/λ. With hc = 1240 eV·nm, a 400 nm photon carries 3.1 eV. Working in electron-volts and nanometres avoids most of the arithmetic slips in this chapter.

The photoelectric effect

Electrons are emitted only above the threshold frequency, whatever the intensity. K_max = hf − φ, and the stopping potential satisfies eV₀ = hf − φ. Intensity controls how many electrons come out — the saturation current — not how fast they travel.

Matter waves

Every particle has λ = h/p = h/√(2mK). For an electron accelerated through a potential V, this reduces to λ ≈ 12.27/√V in ångströms with V in volts, which is the form the exam usually wants.

The Bohr model

For a hydrogen-like atom, r_n = 0.529 n²/Z Å and E_n = −13.6 Z²/n² eV. Energy is negative because the electron is bound; the difference between two levels is what appears as a photon.

Spectral series

The emitted wavelength follows 1/λ = RZ²(1/n₁² − 1/n₂²) with n₁ < n₂. Lyman ends at n = 1 and lies in the ultraviolet; Balmer ends at n = 2 and gives the visible lines, of which n = 3 → 2 is the familiar 656 nm red.

Nuclei, binding energy and decay

Mass defect converts to binding energy through E = Δm c², with 1 u = 931.5 MeV. Binding energy per nucleon peaks near A = 56, which is why both fusion of light nuclei and fission of heavy ones release energy. Decay is exponential: N = N₀e^(−λt).

Photon energy

E = hf = hc/λ · hc = 1240 eV·nm

λ in nm gives E directly in eV.

Photoelectric equation

K_max = hf − φ · eV₀ = hf − φ

V₀ is the stopping potential.

Threshold wavelength

λ₀ = hc/φ

No emission for λ > λ₀, however intense the light.

de Broglie wavelength

λ = h/p = h/√(2mK) · λ(Å) ≈ 12.27/√V

The second form is for an electron accelerated through V volts.

Bohr radius and energy

r_n = 0.529 n²/Z Å · E_n = −13.6 Z²/n² eV

Hydrogen-like species only (one electron).

Rydberg formula

1/λ = R Z² (1/n₁² − 1/n₂²) · R = 1.097 × 10⁷ m⁻¹

n₁ is the lower level.

Radioactive decay

N = N₀ e^(−λt) · A = λN

Activity falls with the same exponential as the number of nuclei.

Half-life and mean life

t½ = 0.693/λ · τ = 1/λ = t½/0.693

After n half-lives, the fraction left is (1/2)ⁿ.

Mass–energy equivalence

E = Δm c² · 1 u = 931.5 MeV/c²

Mass defect Δm gives the binding energy.

Nuclear radius

R = R₀ A^(1/3) · R₀ ≈ 1.2 fm

Density is therefore nearly the same for all nuclei.

Photoelectric effect · 3 marks · easy

The threshold wavelength for a metal is 600 nm. Light of wavelength 400 nm falls on it. Find the work function of the metal and the stopping potential.

1. The work function follows from the threshold wavelength: φ = hc/λ₀ = 1240/600 = 2.07 eV.
2. The incident photon carries E = hc/λ = 1240/400 = 3.10 eV.
3. Maximum kinetic energy: K_max = E − φ = 3.10 − 2.07 = 1.03 eV.
4. The stopping potential is numerically equal to K_max in electron-volts: V₀ = 1.03 V.
5. Sanity check: the incident wavelength is shorter than the threshold, so emission does occur.

Answer: φ ≈ 2.07 eV; V₀ ≈ 1.03 V.

Bohr model and spectra · 3 marks · easy

An electron in a hydrogen atom falls from n = 3 to n = 2. Find the energy of the emitted photon and its wavelength.

1. E₃ = −13.6/9 = −1.51 eV and E₂ = −13.6/4 = −3.40 eV.
2. Photon energy = E₃ − E₂ = −1.51 − (−3.40) = 1.89 eV.
3. Wavelength: λ = 1240/1.89 ≈ 656 nm.
4. This is the red Hα line of the Balmer series, which is why it lies in the visible range.

Answer: E ≈ 1.89 eV; λ ≈ 656 nm (visible red).

Nuclear decay · 3 marks · moderate

A radioactive nuclide has a half-life of 8 hours. What fraction of an initial sample remains after 24 hours, and what is its decay constant?

1. 24 hours is 24/8 = 3 half-lives.
2. Fraction remaining = (1/2)³ = 1/8 = 12.5%.
3. Decay constant: λ = 0.693/t½ = 0.693/8 ≈ 0.0866 per hour.
4. In SI: λ = 0.0866/3600 ≈ 2.4 × 10⁻⁵ s⁻¹.

Answer: One-eighth of the sample remains; λ ≈ 0.087 h⁻¹ (2.4 × 10⁻⁵ s⁻¹).