JEE Physics - MCQ Practice Questions

Alpha decay: A decreases by 4, Z decreases by 2. ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He. The resulting nucleus is Lead-206.

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

Bremsstrahlung (braking radiation) occurs when fast electrons are decelerated by the nucleus, producing continuous X-ray spectrum.

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

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

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

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

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

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

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

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

11,460 years = 2 × 5730 years = 2 half-lives. After 2 half-lives: 1 × ($\frac{1}{2}$)² = 0.25 mg remains.

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.

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.