JEE Physics - MCQ Practice Questions

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

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

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.

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.

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

Using λ = h/√(2mKE), where m = 1.67×10⁻²⁷ kg, KE = 1.6×10⁻¹⁹ J, h = 6.63×10⁻³⁴ J·s. Calculation yields λ ≈ 0.286 nm.

Half-life t₁/₂ = ln(2)/λ. Since λₐ = 2λᵦ, we have t₁/₂ₐ/t₁/₂ᵦ = λᵦ/λₐ = $\frac{1}{2}$. Therefore, t₁/₂ₐ : t₁/₂ᵦ = 1:2.

Using 1/λ = R(1/n₁² - 1/n₂²). For 3→1: 1/λ₃₋₁ = R(1 - $\frac{1}{9}$) = 8R/9. For Lyman alpha 2→1: 1/λ₂₋₁ = R(1 - $\frac{1}{4}$) = 3R/4. Ratio λ₃₋₁/λ₂₋₁ = ($\frac{3}{4}$)/($\frac{8}{9}$) = $\frac{27}{32}$. So λ₂₋₁/λ₃₋₁ = $\frac{32}{27}$.

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).

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

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

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.

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.