A) 25
B) 50
C) 75
D) 80
Correct Answer: A
Solution :
[a] Energy of on photon |
\[=\frac{12400}{3100}=4\,eV=\]\[4\times 96=384\,\,kJ\,mo{{l}^{-1}}\] |
\[\therefore \] % of energy converted to |
\[K.E.=\frac{384-288}{384}=\frac{96}{384}\times 100=25%\] |
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