E=λhc×1000NA
=663×10−9×10006.63×10−34×3×108×6.02×1023
=3×6.02×10kJ
=180.6kJ
JEE Main 2021 — Chemistry Physical Chemistry
Electromagnetic radiation of wavelength 663nm is just sufficient to ionise the atom of metal A. The ionization energy of metal A in kJmol−1 is ___. (Rounded-off to the nearest integer) [h=6.63×10−34Js,c=3.00×108ms−1,NA=6.02×1023mol−1]
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