The K α X-ray of molybdenum has wavelength 0.071 nm. If the energy of a molybdenum atoms 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]

Solution

Physics sir · 4 steps

The K-alpha photon carries away exactly the energy the atom loses when its hole moves from K to L.

1. What K-alpha means. An atom with a K vacancy has energy E_K = 27.5 keV. An L electron drops into that vacancy, so the hole is now in the L shell, and the atom has energy E_L. The difference leaves as one K-alpha photon.

   E_K - E_L = hc/λ

2. Photon energy. hc/λ = (4.14 × 10⁻¹⁵ × 3 × 10⁸)/(0.071 × 10⁻⁹) eV = (1.242 × 10⁻⁶)/(7.1 × 10⁻¹¹) eV ≈ 17 493 eV = 17.49 keV.
3. Subtract. E_L = 27.5 - 17.49 = 10.01 keV.
4. So. The atom with an L electron knocked out has energy of about 10 keV.

The common slip: 17. Reporting the K-alpha photon energy itself gives 17, not 10. The photon energy is the gap between the two states, so subtract it from 27.5 keV.

The idea to keep. Think of X-ray lines as the hole jumping outward: the atom's energy falls from the K-hole state to the L-hole state, and the photon carries away exactly that difference.

Answer from the NTA final answer key. Solution by PrepWiser.