JEE Physics - MCQ Practice Questions

In Bohr's model, angular momentum is quantized as L = mₑvr = nℏ, where n = 1, 2, 3... and ℏ = h/2π.

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.

The 2.22 MeV is the binding energy. Threshold photon energy is slightly higher (≈2.24 MeV) to account for recoil of products.

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).

Conservation of mass number: 14 + 4 = 17 + A → A = 1. Conservation of atomic number: 7 + 2 = 8 + Z → Z = 1. But Z = 0 for neutron. Correction: 7 + 2 = 8 + Z → Z = 1, which is a proton. But mass number gives 18 = 17 + 1. Need neutron.

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

Nuclear forces are strong forces that are attractive at normal nuclear distances (~1 fm) but become repulsive at very short distances (<0.5 fm), creating a potential well. They act between all nucleons (protons and neutrons).

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 ($\frac{1}{4}$ - $\frac{1}{16}$) = ($\frac{1}{9}$ - $\frac{1}{25}$), but this is not true. Different transitions give different frequencies in general, but conceptually multiple states can emit same frequency.

For hydrogen-like ions: IE = 13.6 × Z² eV. For He⁺, Z = 2, so IE = 13.6 × 4 = 54.4 eV.

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

The ionization energy of hydrogen in ground state is 13.6 eV. A photon of exactly 13.6 eV can just ionize the atom with zero kinetic energy of the ejected electron.

The radius of nth orbit in hydrogen atom is given by rₙ = n²a₀/Z, where a₀ = 0.53 Å is the Bohr radius and Z is atomic number.

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.

In β⁻ decay, a neutron converts to proton, emitting electron and antineutrino: n → p + e⁻ + ν̄ₑ

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.

Alpha particles (⁴₂He) are helium nuclei containing 2 protons and 2 neutrons. They have low penetrating power but high ionizing power.

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.