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The dissociation energy of H(2) is 430.5...

The dissociation energy of `H_(2)` is `430.53kJ//mol`.If `H_(2)` is exposed to radiant energy of wavelength `253.7nm`.What `%` of radiant energy will be converted into K.E?

A

0.0886

B

0.0233

C

`1.3%`

D

`90%`

Text Solution

Verified by Experts

The correct Answer is:
A
`(hc)/(lamda)=(430.53xx10^(3))/(6.023xx10^(23))+K.E.`
`K.E =(6.626xx10^(-34)xx3xx10^(8))/(253.7xx10^(-9))-(430.53xx10^(3))/(6.023xx10^(23))`
`=6.9xx10^(-20)`
`:.` Fraction `=(6.9xx10^(-20))/(7.86xx10^(-19))=0.088`
`=8.86%`
Hence a is the correct answer.
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