JEE Physics - MCQ Practice Questions

Using E=mc², the rest mass energy of electron (m=9.11×10⁻³¹ kg) is 0.511 MeV, a fundamental constant in modern physics.

Using hf = Φ + eV₀, when frequency doubles: h(2f) = Φ + eV'₀. Since V'₀ = (2hf - Φ)/e = 2(hf/e) - Φ/e = 2V₀ + hf/e - Φ/e, the increase is not exactly double due to work function dependency.

de Broglie proposed that λ = h/p, where h is Planck's constant and p is momentum. This is fundamental to wave-particle duality.

Using f = cR(1/n₁² - 1/n₂²) = 3×10⁸ × 1.097×10⁷ × ($\frac{1}{1}$ - $\frac{1}{9}$) = 3×10⁸ × 1.097×10⁷ × ($\frac{8}{9}$) ≈ 2.93×10¹⁵ Hz

Iron-56 has the maximum binding energy per nucleon (~8.8 MeV/nucleon), making it the most stable nucleus. This is a key concept in nuclear physics.

Number of half-lives = $\frac{30}{10}$ = 3. Remaining fraction = ($\frac{1}{2}$)³ = $\frac{1}{8}$

At threshold: hf₀ = Φ. For frequency 2f₀: KEₘₐₓ = h(2f₀) - Φ = 2hf₀ - hf₀ = hf₀

Compton scattering shows that photons behave as particles with momentum, providing evidence for particle nature of light. Wavelength shift Δλ = (h/mₑc)(1 - cosθ).

Since c = fλ, we have f = c/λ. Therefore E = hf = hc/λ

λ = h/√(2mKE). Since KE ∝ V, λ ∝ 1/√V. Therefore λ₁/λ₂ = √($\frac{400}{100}$) = √4 = 2

Activity A = λN. When activity decreases by 50% in 1 hour, this means N (and A) reduced to half in 1 hour, which is exactly the definition of half-life.

Threshold wavelength λ₀ = hc/Φ = (1240 eVnm)/(2.3 eV) ≈ 539 nm. Photoelectric effect occurs for λ < 539 nm. Only 300 nm satisfies this.

rₙ = n²a₀, where a₀ is Bohr radius (0.53 Å). Orbital radius is proportional to n².

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)

For hydrogen-like atoms: E = -13.6Z²/n² eV. For He⁺ (Z=2): E = -13.6×4 = -54.4 eV. For H (Z=1): E = -13.6 eV. Ratio is 4:1

Characteristic X-rays arise from inner-shell electron transitions (e.g., L→K shell). Each element has specific wavelengths, hence 'characteristic'.

ΔxΔp ≥ h/4π. If Δx = 0, then Δp must be ≥ ∞ to satisfy the inequality, making momentum completely uncertain.

λ = h/√(2mKE). For same KE: λₑ/λₚ = √(mₚ/mₑ) ≈ √1836 ≈ 42.8

Pair production creates an electron-positron pair. Minimum energy = 2mₑc² = 2 × 0.511 MeV = 1.022 MeV

α-particle: charge = 2e, mass = 4u. Proton: charge = e, mass = u. (e/m)ₐ/(e/m)ₚ = (2e/4u)/(e/u) = $\frac{1}{2}$