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The energy of one mole of photons of radiation of wavelength 300 nm is

(Given : \mathrm{h=6.63 \times 10^{-34}Js, \;\;N_{A}=6.02 \times 10^{23}mol^{-1},\;\;c=3 \times 10^{8}ms^{-1}})

Option: 1

\mathrm{235 \;kJ\; mol^{-1}}


Option: 2

\mathrm{325 \;kJ\; mol^{-1}}


Option: 3

\mathrm{399 \;kJ\; mol^{-1}}


Option: 4

\mathrm{435 \;kJ\; mol^{-1}}


Answers (1)

best_answer

We know,

\text { Energy of one photon, } \mathrm{E=h \nu=\frac{h c}{\lambda}}

\text{Energy of 1 mole photon} =\mathrm{E \times N_{A}=\frac{hc}{\lambda}\times NA}

                                                       \begin{aligned} &=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-2}} \\ &=399 \mathrm{~kJ} \mathrm{~mol}^{-2} \end{aligned}

Hence, the correct answer is Option (3)

Posted by

Ritika Kankaria

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