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JEE Physics
Modern Physics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

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Topics in JEE Physics
Q.1 Medium Modern Physics
A photon of wavelength 100 pm strikes a stationary free electron. After Compton scattering at an angle of 60°, the wavelength of the scattered photon is found to be 102.4 pm. What is the kinetic energy of the recoil electron? (Given: h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s, mₑ = 9.1 × 10⁻³¹ kg)
A 1.65 keV
B 2.43 keV
C 3.21 keV
D 0.98 keV
Correct Answer:  A. 1.65 keV
EXPLANATION

Using Compton scattering formula: λ' - λ = (h/mₑc)(1 - cos θ). With given values, the wavelength shift is 2.4 pm. Using energy conservation, the incident photon energy is E₀ = hc/λ ≈ 12.4 keV. The scattered photon energy E' = hc/λ' ≈ 12.1 keV. The kinetic energy of electron = E₀ - E' ≈ 1.65 keV.

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Q.2 Hard Modern Physics
For a nucleus, the neutron-to-proton ratio (N/Z) increases with mass number. This is because:
A Coulomb repulsion increases faster than strong nuclear force
B Neutrons are heavier than protons
C Extra neutrons increase binding energy more efficiently for heavy nuclei
D Pauli exclusion principle requires more neutrons
Correct Answer:  C. Extra neutrons increase binding energy more efficiently for heavy nuclei
EXPLANATION

For heavy nuclei, the Coulomb repulsion between protons increases significantly. Extra neutrons (uncharged) help stabilize the nucleus without increasing repulsion, requiring N > Z for stability.

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Q.3 Easy Modern Physics
The half-life of ¹⁴C is 5730 years. A sample contains 1 mg of ¹⁴C. After 11,460 years, the remaining mass will be:
A 0.25 mg
B 0.5 mg
C 0.125 mg
D 1 mg
Correct Answer:  A. 0.25 mg
EXPLANATION

11,460 years = 2 × 5730 years = 2 half-lives. After 2 half-lives: 1 × (1/2)² = 0.25 mg remains.

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Q.4 Hard Modern Physics
In pair production, a photon with energy 3 MeV converts near a nucleus into an electron-positron pair. The rest mass energy of electron/positron is 0.51 MeV. The excess energy appears as:
A Kinetic energy of electron and positron
B Heat in the nucleus
C Additional photons
D Recoil energy of nucleus
Correct Answer:  A. Kinetic energy of electron and positron
EXPLANATION

In pair production: E_photon = 2m_e c² + KE_total. Excess = 3 - 2(0.51) = 1.98 MeV becomes kinetic energy of the pair.

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Q.5 Hard Modern Physics
An electron transitions from n=3 to n=1 in a hydrogen atom. How many distinct spectral lines can be observed from all possible transitions?
A 1 line
B 2 lines
C 3 lines
D 4 lines
Correct Answer:  C. 3 lines
EXPLANATION

Possible transitions: 3→1 (direct), 3→2, 2→1. Total = 3 distinct lines. The electron can go 3→2→1 or 3→1 directly.

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Q.6 Medium Modern Physics
The intensity of characteristic X-rays depends on:
A Only the atomic number of the target material
B Only the kinetic energy of incident electrons
C Both atomic number and electron beam current
D The frequency of X-rays only
Correct Answer:  C. Both atomic number and electron beam current
EXPLANATION

Characteristic X-ray intensity depends on the number of inner-shell electrons available (atomic number) and the number of incident electrons (beam current).

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Q.7 Hard Modern Physics
When ²³⁵U nucleus undergoes fission, the total mass defect is 0.1 u. The energy released is approximately:
A 93 MeV
B 140 MeV
C 200 MeV
D 186 MeV
Correct Answer:  A. 93 MeV
EXPLANATION

E = Δm·c² = 0.1 × 931 MeV ≈ 93 MeV. (Note: typical U-235 fission releases ~200 MeV total, distributed among products).

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Q.8 Medium Modern Physics
According to the uncertainty principle, if the position of an electron is known with uncertainty Δx = 10⁻¹⁰ m, the minimum uncertainty in momentum is:
A 5.27 × 10⁻²⁵ kg·m/s
B 1.055 × 10⁻²⁴ kg·m/s
C 3.16 × 10⁻²⁴ kg·m/s
D 2.64 × 10⁻²⁵ kg·m/s
Correct Answer:  A. 5.27 × 10⁻²⁵ kg·m/s
EXPLANATION

Δx·Δp ≥ h/(4π). Δp ≥ 1.055×10⁻³⁴/(4π×10⁻¹⁰) ≈ 5.27 × 10⁻²⁵ kg·m/s.

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Q.9 Medium Modern Physics
The threshold frequency for photoelectric effect in a metal is 6 × 10¹⁴ Hz. The work function of the metal is:
A 2.48 eV
B 3.97 eV
C 4.14 eV
D 2.0 eV
Correct Answer:  A. 2.48 eV
EXPLANATION

Work function W = hf₀ = 6.63 × 10⁻³⁴ × 6 × 10¹⁴ = 3.98 × 10⁻¹⁹ J ≈ 2.48 eV.

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Q.10 Medium Modern Physics
A radioactive nucleus ₆₀₂₇Co undergoes beta-plus decay. The daughter nucleus is:
A ⁶⁰₂₆Ni
B ⁶⁰₂₈Ni
C ⁶¹₂₇Co
D ⁵⁹₂₆Fe
Correct Answer:  A. ⁶⁰₂₆Ni
EXPLANATION

In beta-plus decay, Z decreases by 1, A remains constant. ⁶⁰₂₇Co → ⁶⁰₂₆Ni + e⁺ + νₑ.

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