JEE Physics - MCQ Practice Questions

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.

The stopping potential is related to maximum kinetic energy by: KE_max = eV_s. For V_s = 2V, KE_max = 2 eV.

In alpha decay, a nucleus emits ⁴₂He. Parent nucleus (A,Z) → Daughter (A-4, Z-2) + Alpha particle (4,2).

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

Activity A = dN/dt = λN, measured in Becquerels (decays per second) or Curies. It represents the decay rate.

Angular momentum L = ℏ√(l(l+1)) = 2ℏ gives l(l+1) = 4, so l = 2. Since l < n, minimum n = 3.

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

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).

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

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.

²₁H + ²₁H → ⁴₂He + energy. Using mass defect and E=mc²: Energy released ≈ 23.8 MeV. This is the basis of thermonuclear fusion.

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

From Einstein's photoelectric equation: KE_max = hf - φ. As frequency increases, maximum kinetic energy increases proportionally. Stopping potential depends on frequency, KE depends on frequency (not intensity), and threshold frequency is directly related to work function.

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.