JEE Physics - MCQ Practice Questions

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.

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

In beta-minus decay, a neutron converts to a proton, emitting an electron and antineutrino. Atomic number increases by 1: Z = 6 + 1 = 7. The daughter nucleus is ¹⁴₇N.

After 30 days = 3 half-lives. Remaining fraction = ($\frac{1}{2}$)³ = $\frac{1}{8}$

Bohr's quantization condition: L = mvr = n(h/2π) = nℏ, where n = 1, 2, 3...

ν = Rc(1/n₁² - 1/n₂²) = 1.097×10⁷ × 3×10⁸ × (1 - $\frac{1}{9}$) = 1.097×10¹⁵ × $\frac{8}{9}$ ≈ 2.47×10¹⁵ Hz

R = 1.2 × (88)^($\frac{1}{3}$) ≈ 1.2 × 3.65 ≈ 4.4 fm

From N = N₀e^(-λt), at t = t₁/₂, N = N₀/2, so $\frac{1}{2}$ = e^(-λt₁/₂), giving λ = ln2/t₁/₂