Electromagnetic radiation of wavelength 663 nm is just sufficient to ionize the atom of metal A. The ionization energy of metal A in kJ mol-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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answer is 181.

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Detailed Solution

I.E=hcλ=6.63×10-34×3×108663×10-9×6.02×1023=180.6KJ/mole=181KJ

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