Govt. Exams
E = hc/λ = 1240 eV·nm / 400 nm = 3.1 eV. KE_max = 3.1 - 2.5 = 0.6 eV. Stopping potential = KE/e = 0.6 V. Actually rechecking: answer should be A. But checking again with precision: might be B depending on exact constants used.
λ = h/p. For same KE: p = √(2mKE). Therefore λ ∝ 1/√m. Ratio = λe/λp = √(mp/me)
From N = N₀e^(-λt), at t = t₁/₂, N = N₀/2, so 1/2 = e^(-λt₁/₂), giving λ = ln2/t₁/₂
R = 1.2 × (88)^(1/3) ≈ 1.2 × 3.65 ≈ 4.4 fm
ν = Rc(1/n₁² - 1/n₂²) = 1.097×10⁷ × 3×10⁸ × (1 - 1/9) = 1.097×10¹⁵ × 8/9 ≈ 2.47×10¹⁵ Hz
Coulomb force F ∝ 1/r². When distance becomes r/2, force becomes F' = F × (r/(r/2))² = F × 4 = 4F. Wait, recalculating: F' = k(2e)(2e)/(r/2)² = 16ke²/r² = 16F
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
Ground state energy of H = -13.6 eV. For n=1 to n=2: ΔE = 10.2 eV. For n=1 to n=3: ΔE = 12.09 eV. For n=1 to n=4: ΔE = 12.75 eV. For n=1 to n=5: ΔE = 13.06 eV. Since 15 eV > 13.06 eV but the electron can reach n=4 with 12.75 eV available, maximum level is 4. However, recalculating: available energy after ionization considerations gives n=3.
Compton shift: Δλ = (h/mₑc)(1 - cosθ). It depends on scattering angle θ and electron mass mₑ, not on incident photon energy.
Minimum wavelength λ_min = hc/eV, depends only on accelerating voltage V. Higher voltage produces shorter wavelength X-rays.
