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Modern Physics: concepts, formulas & solved PYQ-style examples

Modern physics is short, formula-driven and one of the best chapters to score in quickly. Photoelectric effect and the Bohr model are asked regularly.

Work in electron-volts: hc ≈ 1240 eV·nm turns most photon-energy calculations into one division.

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Key concepts

Photoelectric effect

Einstein’s equation; stopping potential depends on frequency, current on intensity.

Matter waves

de Broglie wavelength λ = h/p for electrons accelerated through a potential V.

Bohr model

Radius ∝ n²/Z, energy ∝ −Z²/n², spectral series of hydrogen.

X-rays

Cut-off wavelength depends only on accelerating voltage.

Nuclei and radioactivity

Binding energy, half-life and the decay law.

Formula sheet

Photon energy

E=hν=hcλ≈1240λ (nm) eVE = h\nu = \dfrac{hc}{\lambda} \approx \dfrac{1240}{\lambda\,(\text{nm})}\ \text{eV}

Einstein’s equation

Kmax=hν−ϕ=eV0K_{max} = h\nu - \phi = eV_0

de Broglie wavelength

λ=hp=h2meV\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2meV}}

Bohr energy levels

En=−13.6 Z2n2 eVE_n = -\dfrac{13.6\,Z^2}{n^2}\ \text{eV}

Rydberg formula

1λ=RZ2(1n12−1n22)\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)

Radioactive decay

N=N0e−λt,T1/2=ln⁡2λN = N_0 e^{-\lambda t},\quad T_{1/2} = \dfrac{\ln 2}{\lambda}

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Stopping potential or maximum kinetic energy for a given wavelength and work function.
  • How photoelectric current and stopping potential change with intensity vs frequency.
  • Wavelength of a hydrogen spectral line and identifying its series.
  • Ratio questions: radius, speed or energy of electron orbits for different n and Z.

Solved examples

Example 1

Light of wavelength 300 nm falls on a metal with work function 2.5 eV. Find the maximum kinetic energy of photoelectrons and the stopping potential.

  1. 1Photon energy: E=1240300≈4.13E = \dfrac{1240}{300} \approx 4.13 eV.
  2. 2Einstein’s equation: Kmax=4.13−2.5=1.63K_{max} = 4.13 - 2.5 = 1.63 eV.
  3. 3Stopping potential: eV0=Kmax⇒V0≈1.63eV_0 = K_{max} \Rightarrow V_0 \approx 1.63 V.
Answer: Kmax≈1.63K_{max} \approx 1.63 eV, V0≈1.63V_0 \approx 1.63 V
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Example 2

Find the wavelength of light emitted when an electron in hydrogen falls from n = 3 to n = 2 (R = 1.097 × 10⁷ m⁻¹).

  1. 1Rydberg formula: 1λ=R(122−132)=5R36\dfrac{1}{\lambda} = R\left(\dfrac{1}{2^2} - \dfrac{1}{3^2}\right) = \dfrac{5R}{36}.
  2. 2λ=365×1.097×107≈6.56×10−7\lambda = \dfrac{36}{5 \times 1.097\times10^{7}} \approx 6.56\times10^{-7} m.
  3. 3Lines ending at n = 2 belong to the Balmer series (visible light).
Answer: λ≈656\lambda \approx 656 nm (red Balmer line)
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Common traps

  • Thinking brighter light gives faster photoelectrons — intensity changes the current, not K_max.
  • Forgetting Z² for hydrogen-like ions such as He⁺ and Li²⁺.
  • Mixing joules and electron-volts in the same equation.

Practise with

ElectrostaticsRay OpticsCurrent ElectricityHydrogen spectrum (Bohr model) simulationPhotoelectric effect and matter waves simulationAtoms: Rutherford, Bohr and X-rays simulationNuclear physics: binding energy and radioactivity simulation239 Atoms & Nuclei practice questions200 Dual Nature of Matter practice questions70 Experimental Physics practice questions5 Modern Physics practice questionsVideo solutions

FAQ

Does stopping potential depend on light intensity?

No. Stopping potential depends only on the frequency of light and the work function; intensity changes the photoelectric current.

Why use 1240 in photon-energy questions?

hc ≈ 1240 eV·nm, so E (eV) = 1240 / λ (nm) — a fast, exam-friendly shortcut.

Is modern physics high-weightage in NEET?

It is one of the most scoring chapters for NEET because the questions are short and formula-based.

Stuck on a Modern Physics question?

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