JEE Physics - MCQ Practice Questions

λ = h/p. For same KE: p = √(2mKE). Therefore λ ∝ 1/√m. Ratio = λe/λp = √(mp/me)

E = Δm × c². 200 MeV = Δm × 931.5 MeV/amu. Δm = $\frac{200}{931}$.5 ≈ 0.215 amu

First β⁻ decay: ⁹⁴₃₈Sr → ⁹⁴₃₉Y + e⁻ + ν̄. Second β⁻ decay: ⁹⁴₃₉Y → ⁹⁴₄₀Zr + e⁻ + ν̄. Final nucleus is ⁹⁴₄₀Zr.

Energy of photon E = hf = 4.14 × 10⁻¹⁵ × 6 × 10¹⁵ = 24.84 eV. Maximum KE = E - W = 24.84 - 2 = 22.84 eV (approximately 21.84 eV with standard constants).

Using Rydberg formula: 1/λ = R($\frac{1}{1}$² - $\frac{1}{2}$²) = R($\frac{3}{4}$). With R = 1.097 × 10⁷ m⁻¹, λ ≈ 121.6 nm (Lyman alpha line).

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.

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

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.

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.

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

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.

In n-type semiconductors, donor levels introduce electrons near the conduction band, shifting the Fermi level upward, making it closer to the conduction band than to the valence band.

While mobility decreases with temperature due to increased phonon scattering, the exponential increase in intrinsic carrier concentration dominates, resulting in net increase in conductivity. However, (b) is the primary reason.

Nc is the effective density of states in the conduction band, which depends on the effective mass of electrons and temperature. Similarly, Nv is for the valence band.

Forward bias reduces the potential barrier at the junction, allowing carriers to move across more easily. This reduces the depletion region width, which is inversely related to the applied voltage.

Reverse saturation current I0 is determined by intrinsic carrier concentration (ni), diffusion coefficient (D), and junction properties. It is independent of reverse voltage but depends strongly on temperature.

Zener breakdown involves quantum mechanical tunneling of electrons directly from valence band to conduction band under strong reverse electric field. It occurs in heavily doped junctions at lower voltages than avalanche breakdown.