JEE Physics - MCQ Practice Questions

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

In β⁻ decay, a neutron converts to proton, but this occurs in daughter nucleus. Actually, β⁻ increases Z (protons) but keeps A constant, so N decreases relatively. In β⁺ decay, proton decreases. Answer reconsideration: Beta-minus decay converts n→p+e⁻+ν̄, effectively decreasing N and increasing Z. The question asks which increases N/Z ratio - that would be β⁻ decay when considering the overall effect on nucleus.

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

Δx·Δp ≥ h/4π. Δv ≥ h/(4πm·Δx) = 6.63×10⁻³⁴/(4π×9.1×10⁻³¹×0.1×10⁻⁹) ≈ 5.8×10⁶ m/s

When fast electrons knock out inner shell (K-shell) electrons, electrons from outer shells fall down, emitting characteristic X-rays specific to the target material.

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).

Stopping potential Vs = (hf - W)/e, which depends directly on frequency. Intensity affects photocurrent, not maximum KE. Threshold frequency exists. Emission requires photon energy > work function.

Bohr's quantization condition states that angular momentum L = mvr = nh/(2π), where n is the principal quantum number.

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).

Higher binding energy per nucleon indicates greater nuclear stability. ⁵⁶Fe (8.79 MeV/nucleon) is more stable than ²³⁵U (7.59 MeV/nucleon).

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.