The \( K_\alpha \) X-ray of molybdenum has wavelength \( 0.071 \, \text{nm} \). If the energy of a molybdenum atom with a K electron knocked out is \( 27.5 \, \text{keV} \), the energy of this atom when an L electron is knocked out will be ______ \( \text{keV} \). (Round off to the nearest integer)
[\( h = 4.14 \times 10^{-15} \, \text{eVs} \), \( c = 3 \times 10^{8} \, \text{ms}^{-1} \)]