The accelerating potential in a Coolidge X-ray tube is increased from 20 kV to 30 kV. If the percentage change in its minimum emitted wavelength is (10x + 3.3)%, find x.
Physics · Mixed Chapter
JEE Mains

Question

The accelerating potential in a Coolidge X-ray tube is increased from 20 kV to 30 kV. If the percentage change in its minimum emitted wavelength is (10x + 3.3)%, find x.

G1 Physics · Accepted Answer

Step 1 — Duane–Hunt law states that the shortest X-ray wavelength is λ₀ = hc/(eV). Step 2 — Initially V₁ = 20 kV and finally V₂ = 30 kV. Consequently, λ₂/λ₁ = V₁/V₂ = 20/30 = 2/3. Step 3 — Percentage reduction in λ is 100 × (λ₁ − λ₂)/λ₁ = 100 × (1 − 2/3) = 33.33%. Step 4 — From the question, the percentage change = (10x + 3.3)%. Thus 10x + 3.3 ≈ 33.3 ⇒ 10x ≈ 30 ⇒ x = 3. Answer: 3.