JEE Physics - MCQ Practice Questions

Compton scattering shows that photons behave as particles with momentum, providing evidence for particle nature of light. Wavelength shift Δλ = (h/mₑc)(1 - cosθ).

Since c = fλ, we have f = c/λ. Therefore E = hf = hc/λ

λ = h/√(2mKE). Since KE ∝ V, λ ∝ 1/√V. Therefore λ₁/λ₂ = √($\frac{400}{100}$) = √4 = 2

Activity A = λN. When activity decreases by 50% in 1 hour, this means N (and A) reduced to half in 1 hour, which is exactly the definition of half-life.

Threshold wavelength λ₀ = hc/Φ = (1240 eVnm)/(2.3 eV) ≈ 539 nm. Photoelectric effect occurs for λ < 539 nm. Only 300 nm satisfies this.

rₙ = n²a₀, where a₀ is Bohr radius (0.53 Å). Orbital radius is proportional to n².

N = (1g/100g·mol⁻¹) × Nₐ = 6.02×10²¹. λ = A/N = 10¹⁵/(6.02×10²¹) ≈ 1.66×10⁻⁷ s⁻¹. (Re-check: should be 10¹⁵/6.02×10²¹ ≈ 1.66×10⁻⁷)

For hydrogen-like atoms: E = -13.6Z²/n² eV. For He⁺ (Z=2): E = -13.6×4 = -54.4 eV. For H (Z=1): E = -13.6 eV. Ratio is 4:1

Characteristic X-rays arise from inner-shell electron transitions (e.g., L→K shell). Each element has specific wavelengths, hence 'characteristic'.

ΔxΔp ≥ h/4π. If Δx = 0, then Δp must be ≥ ∞ to satisfy the inequality, making momentum completely uncertain.

λ = h/√(2mKE). For same KE: λₑ/λₚ = √(mₚ/mₑ) ≈ √1836 ≈ 42.8

Pair production creates an electron-positron pair. Minimum energy = 2mₑc² = 2 × 0.511 MeV = 1.022 MeV

α-particle: charge = 2e, mass = 4u. Proton: charge = e, mass = u. (e/m)ₐ/(e/m)ₚ = (2e/4u)/(e/u) = $\frac{1}{2}$

Ionization energy of hydrogen = 13.6 eV. A photon with exactly 13.6 eV provides just enough energy to ionize, leaving zero kinetic energy for the electron.

Gamma rays have the highest penetrating power due to their high energy and lack of charge. Beta particles have moderate penetrating power. Alpha particles have the lowest penetrating power.

Using λ = h/√(2mKE), where m = 1.67×10⁻²⁷ kg, KE = 1.6×10⁻¹⁹ J, h = 6.63×10⁻³⁴ J·s. Calculation yields λ ≈ 0.286 nm.

Half-life t₁/₂ = ln(2)/λ. Since λₐ = 2λᵦ, we have t₁/₂ₐ/t₁/₂ᵦ = λᵦ/λₐ = $\frac{1}{2}$. Therefore, t₁/₂ₐ : t₁/₂ᵦ = 1:2.

Total binding energy = (Binding energy per nucleon) × (Mass number) = 8.8 × 256 ≈ 2253 MeV ≈ 2250 MeV.

In beta-minus decay, a neutron converts to a proton, so atomic number (Z) increases by 1. Mass number (A) remains unchanged.

Using 1/λ = R(1/n₁² - 1/n₂²). For 3→1: 1/λ₃₋₁ = R(1 - $\frac{1}{9}$) = 8R/9. For Lyman alpha 2→1: 1/λ₂₋₁ = R(1 - $\frac{1}{4}$) = 3R/4. Ratio λ₃₋₁/λ₂₋₁ = ($\frac{3}{4}$)/($\frac{8}{9}$) = $\frac{27}{32}$. So λ₂₋₁/λ₃₋₁ = $\frac{32}{27}$.