If the lowest energy X-rays have λ=3.055×10−8 m, estimate the minimum difference in energy between two Bohr's orbits such that an electronic transition would correspond to the emission of an X-ray. Assuming that the electrons in other shells exert no influence, at what Z(minimum) would a transition from a second energy level to the first, result in the emission of an X-ray?

Explanation

Part 1: Minimum Energy DifferenceEnergy of an X-ray Photon:The energy of a photon is related to its wavelength by the equation:E=λhc Where:h = Planck's constant (6.626 × 10⁻³⁴ J·s)c = speed of light (3 × 10⁸ m/s)λ = wavelength of the X-ray (3.055 × 10⁻⁸ m)Calculate the Energy:E=6.50×10−18 JConvert to Electron Volts (eV):To convert joules to electron volts, divide by the elementary charge (1.602 × 10⁻¹⁹ C):E(eV)=1.602×10−19 J/eV6.50×10−18 J E(eV)≈40.57 eVTherefore, the minimum difference in energy between two Bohr's orbits is approximately 40.57 eV.Part 2: Minimum Atomic Number (Z)Energy Transition Formula:The energy of an electron in a hydrogen-like atom (with atomic number Z) is given by:En =−13.6n2Z2 eVThe energy difference between two levels (n₂ and n₁) is:ΔE=E2 −E1 =−13.6Z2(n22 1 −n12 1 )Given Transition:We are given that the transition is from n₂ = 2 to n₁ = 1.ΔE=−13.6Z2(221 −121 )ΔE=−13.6Z2(41 −1)ΔE=−13.6Z2(−43 )ΔE=10.2Z2 eVEquate Energy Differences:We know that the energy difference must be at least 40.57 eV for X-ray emission.10.2Z2=40.57Solve for Z:Z2=10.240.57 Z2≈3.977Z≈3.977Z≈1.994Minimum Integer Z:Since Z must be an integer, the minimum atomic number is Z = 2.