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JEE Main & Advanced Chemistry Structure of Atom / परमाणु संरचना Sample Paper Topic Test - Structure of Atom

  • question_answer
    Photon having wavelength 310 nm is used to break the bond of \[{{A}_{2}}\] molecule having bond energy \[288\,kg\,mo{{l}^{-1}}\] then % of energy of photon converted to the K.E. is  \[[hc=12400ev\overset{o}{\mathop{A}}\,,\] \[1ev=96kJ/mol]\]    

    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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