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Questions  >>  CBSE XI  >>  Chemistry  >>  Structure of Atom
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Calculate energy of 2 moles of photons of radiation whose frequency is $5 \times 10^{34}\;Hz$

$\begin{array}{1 1} (a)\;3990.2\;kJmol^{-1} \\ (b)\;5990.2\;kJmol^{-1}\\ (c)\;4990.2\;kJmol^{-1}\\ (d) \;2990.2\;kJmol^{-1} \end{array} $

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A)
Solution :
Energy (E) of one photon $=E=hv$
Where $h= 6.626 \times 10^{-24}\;Js$
$v= 5 \times 1-^{14}s^{-1}$
$E= (6.626 \times 10^{-14}\;s^{-1})$
$\qquad= 3.313 \times 10^{-12}\;J$
Energy of 2 mole of photon $=(3.313 \times 10^{-19}\;J) \times (2 \times 6.022 \times 10^{23}\;mol^{-1})$
$\qquad= 3990.2 \;kJmol^{-1}$
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