The wavelength of the Kα line of silver (Z = 47) is (x/10 + 7/100) × 10⁻¹⁰ m. Given that the screening constant b = 1, find x.
Physics · Mixed Chapter
JEE Mains

Question

The wavelength of the Kα line of silver (Z = 47) is (x/10 + 7/100) × 10⁻¹⁰ m. Given that the screening constant b = 1, find x.

G1 Physics · Accepted Answer

Step 1 — The Kα line corresponds to an electron transition from n = 2 to n = 1. Step 2 — With screening constant b = 1, Moseley’s/Rydberg’s relation is 1/λ = R∞(Z − b)²[(1/1²) − (1/2²)] = (3/4)R∞(Z − b)². Step 3 — For silver Z = 47, so Z − b = 46. Use R∞ = 1.097 × 10⁷ m⁻¹. λ = 4/[3 × (1.097 × 10⁷) × 46²] m = 5.74 × 10⁻¹¹ m. Step 4 — Convert to the form in the question: λ = 0.574 × 10⁻¹⁰ m ≈ 0.57 × 10⁻¹⁰ m. Given λ = (x/10 + 7/100) × 10⁻¹⁰ m. Thus x/10 + 0.07 = 0.57 ⇒ x = 5. Answer: 5.