Atoms And Nuclei

Step 1: Relation of energies. E = -K K = 3.4 eV
Step 2: de-Broglie wavelength. λ = h√(2mK) ≈ 0.66 nm

Step 1: Energy relations. E = (hc)/(λ)
Step 2: Adding transitions. (hc)/(₃) = (hc)/(₁) + (hc)/(₂)
Step 3: Cancelling constants. ₃⁻¹ = ₁⁻¹ + ₂⁻¹

Step 1: Minimum excitation energy of hydrogen:
Step 2: For head-on collision with equal masses, maximum energy transfer is half.
Step 3: Hence neutron must have:

Step 1: For hydrogen-like atoms λ ∝ (1)/(Z²)
Step 2: Atomic numbers H, D: Z=1; He⁺: Z=2; Li²+: Z=3
Step 3: Hence lambda₁=lambda₂=(1)/(1²), lambda₃=(1)/(4), lambda₄=(1)/(9)

Step 1: Energy gained: eV=(1)/(2)mv²
Step 2: v=√((2eV)/(m)) =√frac2(1.6×10⁻19)(3000)9.1×10⁻31 ≈ 3.26×10⁷m/s

Step 1: Balmer third line: (1)/(λ) = RZ²((1)/(2²)-(1)/(5²))
Step 2: Given λ = 108.5nm ⟹ Z=2
Step 3: Ground state energy: E₁ = 13.6 Z² = 54.4eV

Step 1: For excitation of hydrogen atom: Emin = 10.2eV
Step 2: In head-on elastic collision of equal masses, only half energy is transferred. K = 2 × 10.2 = 20.4eV

Step 1: For hydrogen-like atoms: (1)/(λ) ∝ Z² ⟹ λ ∝ (1)/(Z²) Step 2: Nuclear charges: ZH = 1, ZD = 1, ZHe⁺ = 2, ZLᵢ⁺+ = 3 Step 3: Wavelength ratios: lambda₁ : lambda₂ : lambda₃ : lambda₄ = (1)/(1²) : (1)/(1²) : (1)/(2²) : (1)/(3²) = 1 : 1 : (1)/(4) : (1)/(9) ⟹ lambda₁ = lambda₂ = 4lambda₃