The Kα - X ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be _____ keV. (Round off to the nearest integer)
[h = 4.14 × 10⁻¹⁵ eVs, c = 3 × 10⁸ ms⁻¹]
The Kα - X ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atom with a K electron knocked out is 27.5 keV, the energy of this atom when an L electron is knocked out will be _____ keV. (Round off to the nearest integer)
[h = 4.14 × 10⁻¹⁵ eVs, c = 3 × 10⁸ ms⁻¹]
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1 Answer
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E_K - E_L = hc/λ_Kα
E_K - E_L = (4.14×10? ¹? * 3×10? )/0.071×10? = 17500 eV = 17.5 keV.
E_L = E_K - 17.5 = 27.5 - 17.5 = 10 keV.
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