The $K _ { \alpha } \mathrm { X}$-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) $\left[ h = 4.14 \times 10 ^ { - 15 } \mathrm { eV } \mathrm { s } , c = 3 \times 10 ^ { 8 } \mathrm {~m} \mathrm {~s} ^ { - 1 } \right]$