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Vishrant Vasavada Grade: 11
        The following question was asked by me:


The dissociation energy of H2 is 430.53 KJ/mol. If H2 is exposed to radiant energy of wavelength 253.7 nm, what % of radiant energy will be converted into KE?



ask iitian expert ramesh.iitkgp replied as:

The bond dissociation energy for the H-H bond is the energy required to break the H2 molecule (in its most stable geometry) into separate hydrogen atoms.

Radiant energy = 1242 eV-nm / 253.7 nm = 4.895 eV = 7.843x10-19 J. How this Radiant Energy = 1242 eV-nm/253.7 nm came? Explain in detail...
7 years ago

Answers : (2)

Ramesh V
70 Points
										



I assumed you (a JEE prep. student), might be knowing that the product of Plank's constant and velocity of light can be expressed as :                    hc = 1242 eV-nm  and    1 eV = 1.602x10-19 J


and   E = h.c / wavelength


Please check the units..


I hope this makes you clear...


Radiant energy = 1242 eV-nm / 253.7 nm = 4.895 eV = 7.843x10-19 J. How this Radiant Energy = 1242 eV-nm/253.7 nm

7 years ago
Vishrant Vasavada
23 Points
										

While solving numbers of problems, it didn't click in my mind. We were not told about it in coaching class nor I saw in book. Thanks..this formula is of great use!

7 years ago
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