Govt. Exams
For a free electron at rest, if it absorbs a photon with energy E and momentum p=E/c, it cannot emit a photon in any direction while conserving both energy and momentum. This is because the electron would need to have kinetic energy, but no emission direction satisfies both conservation laws simultaneously.
²₁H + ²₁H → ⁴₂He + energy. Using mass defect and E=mc²: Energy released ≈ 23.8 MeV. This is the basis of thermonuclear fusion.
After 2α decays: mass number decreases by 8, atomic number by 4: ²³⁸₉₂U → ²³⁰₈₈Ra. After 2β⁻ decays: atomic number increases by 2: ²³⁰₉₀Th. But rechecking: ²³⁰₈₈Ra is correct intermediate form.
For same KE: λ = h/√(2mKE). λₑ/λₚ = √(mₚ/mₑ) ≈ 42.8 (using mₑ = 9.1×10⁻³¹ kg, mₚ = 1.67×10⁻²⁷ kg).
Angular momentum L = ℏ√(l(l+1)) = 2ℏ gives l(l+1) = 4, so l = 2. Since l < n, minimum n = 3.
Multiple transitions can produce the same photon frequency. For example, n=4→n=2 and n=5→n=3 can produce the same frequency if (1/4 - 1/16) = (1/9 - 1/25), but this is not true. Different transitions give different frequencies in general, but conceptually multiple states can emit same frequency.
The 2.22 MeV is the binding energy. Threshold photon energy is slightly higher (≈2.24 MeV) to account for recoil of products.
By momentum conservation, Pα = Pdaughter. KEα/KEdaughter = mdaughter/mα = (A-4)/4. If KEdaughter = 0.5 MeV, then KEα = 0.5 × 4/(A-4). For typical nuclei (A~200), KEα ≈ 2 × 0.5/(0.8) ≈ 1.25 MeV. Closer approximation gives ~2 MeV.
λ = h/√(2mKE). For same KE: λₑ/λₚ = √(mₚ/mₑ) ≈ √1836 ≈ 42.8
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)
