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JEE Physics
Modern Physics

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

48 Q 9 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 1–10 of 48
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 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.3 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.4 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.5 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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Q.6 Medium Modern Physics
In a nuclear reactor, control rods are used to absorb neutrons. Which isotope is commonly used in control rods?
A Boron-10
B Uranium-235
C Plutonium-239
D Cobalt-60
Correct Answer:  A. Boron-10
EXPLANATION

Boron-10 and Cadmium have high neutron absorption cross-sections and are used in control rods to regulate chain reactions in nuclear reactors.

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Q.7 Medium Modern Physics
The activity of a radioactive sample decreases from 8000 Bq to 1000 Bq in 20 hours. The half-life of the sample is:
A 6.67 hours
B 10 hours
C 13.33 hours
D 20 hours
Correct Answer:  A. 6.67 hours
EXPLANATION

A = A₀(1/2)^(t/T₁/₂). 1000 = 8000(1/2)^(20/T₁/₂). (1/8) = (1/2)³, so 20/T₁/₂ = 3, T₁/₂ = 6.67 hours.

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Q.8 Medium Modern Physics
In Compton scattering, a photon collides with a free electron at rest. Which quantity always increases?
A Photon energy
B Photon wavelength
C Electron rest mass
D Photon frequency
Correct Answer:  B. Photon wavelength
EXPLANATION

In Compton effect, photon transfers energy to electron, losing energy and increasing wavelength. Δλ = (h/m_e c)(1 - cosθ).

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Q.9 Medium Modern Physics
The de Broglie wavelength of an electron with kinetic energy 50 eV is:
A 0.173 nm
B 0.275 nm
C 0.155 nm
D 0.225 nm
Correct Answer:  A. 0.173 nm
EXPLANATION

λ = h/p = h/√(2mE). For 50 eV electron: λ = 6.63×10⁻³⁴/√(2×9.1×10⁻³¹×50×1.6×10⁻¹⁹) ≈ 0.173 nm.

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Q.10 Medium Modern Physics
An electron in the first excited state of hydrogen atom (n=2) transitions to ground state (n=1). The wavelength of emitted photon is approximately:
A 121.6 nm
B 102.6 nm
C 656 nm
D 486 nm
Correct Answer:  A. 121.6 nm
EXPLANATION

Using Rydberg formula: 1/λ = R(1/1² - 1/2²) = R(3/4). With R = 1.097 × 10⁷ m⁻¹, λ ≈ 121.6 nm (Lyman alpha line).

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