A) 100%
B) 8.82%
C) 2.22%
D) 1.22%
Correct Answer: B
Solution :
[b] Energy of 1 mole of photons, \[E={{N}_{0}}\times {{h}_{v}}\] \[=\frac{{{N}_{0}}\times h\times c}{\lambda }\] \[=\frac{6.023\times {{10}^{23}}\times 6.63\times {{10}^{-34}}\times 3\times {{10}^{8}}}{253.7\times {{10}^{-9}}}\] Energy converted into \[KE=\left( 472.2-430.53 \right)kJ%\]% of energy converted into \[KE=\frac{\left( 472.2-430.53 \right)}{472.2}\times 100=8.82%\]%You need to login to perform this action.
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