JEE Physics - MCQ Practice Questions

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.

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 = $\frac{32}{27}$.

Compton scattering formula: Δλ = λ - λ₀ = (h/mₑc)(1 - cos θ). At θ = 90°, Δλ = (h/mₑc)(1 - 0) = h/(mₑc).

If 93.75% decays, 6.25% remains. 6.25% = 6.$\frac{25}{100}$ = $\frac{1}{16}$ = ($\frac{1}{2}$)⁴. So 4 half-lives have passed. Time = 4 × 10 = 40 days. Correction: 6.25% = $\frac{1}{16}$, which requires 4 half-lives = 40 days.

Characteristic X-ray frequency (Moseley's law) depends on atomic number Z of the target. f = R(Z - σ)²(1/n₁² - 1/n₂²). It is independent of incident electron energy (which only affects intensity).

Maximum photon energy = eV. E = hc/λ₀, so λ₀ = hc/eV. This is the minimum wavelength or cutoff wavelength.

Maximum X-ray frequency: fmax = eV/h. If V is doubled, fmax also doubles, since f ∝ V.

Using Rydberg formula: 1/λ = R($\frac{1}{4}$ - $\frac{1}{16}$) = 3R/16. With R = 1.097×10⁷ m⁻¹, we get λ ≈ 486 nm (H-beta line).

Maximum binding energy per nucleon means Fe-56 has highest stability and also highest mass defect. This is the peak of the nuclear stability curve.

Using N = N₀($\frac{1}{2}$)^(t/T₁/₂): 0.75N₀ = N₀($\frac{1}{2}$)^(6/T₁/₂). Taking log: ln(0.75) = (6/T₁/₂)ln(0.5), giving T₁/₂ = 6√3 hours ≈ 10.39 hours.

λ = h/√(2meV). For V = 100V: λ = 12.27/√(2×9.1×10⁻³¹×1.6×10⁻¹⁹×100) ≈ 1.23 Å = 0.123 nm.

Energy at n=3: E₃ = -13.$\frac{6}{9}$ = -1.51 eV. Ionization energy = 0 - (-1.51) = 1.51 eV.

When high-speed electrons are stopped in matter, their kinetic energy produces both continuous X-ray spectrum (Bremsstrahlung radiation) and heat.

Minimum wavelength λ_min = hc/eV, depends only on accelerating voltage V. Higher voltage produces shorter wavelength X-rays.

Compton shift: Δλ = (h/mₑc)(1 - cosθ). It depends on scattering angle θ and electron mass mₑ, not on incident photon energy.

Using λ = h/√(3mkT), where m = 1.67×10⁻²⁷ kg for neutron, h = 6.63×10⁻³⁴ J·s, k = 1.38×10⁻²³ J/K. λ ≈ 0.016 nm

ν = Rc(1/n₁² - 1/n₂²) = 1.097×10⁷ × 3×10⁸ × (1 - $\frac{1}{9}$) = 1.097×10¹⁵ × $\frac{8}{9}$ ≈ 2.47×10¹⁵ Hz

R = 1.2 × (88)^($\frac{1}{3}$) ≈ 1.2 × 3.65 ≈ 4.4 fm

From N = N₀e^(-λt), at t = t₁/₂, N = N₀/2, so $\frac{1}{2}$ = e^(-λt₁/₂), giving λ = ln2/t₁/₂