The K$_\alpha$ 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 = $4.14 \times 10^{-15}$ eVs, c = $3 \times 10^8$ ms$^{-1}$ ]