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ChemistryAtomic StructureMiscellaneous/MixedMedium2 minPYQ_2022

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The energy of one mole of photons of radiation of wavelength $300 nm$ is (Given : $h = 6.63 \times 10^{- 34} J s, N_{A} = 6.02 \times 10^{23} {mol}^{- 1} c = 3 \times 10^{8} m s^{- 1}$ )

Options:

Answer:

Solution:

$λ = 300 \times 10^{- 9} m$

Energy of single photon $= \frac{hc}{λ}$

Energy of $N_{A}$ photons $= \frac{6 .02 \times 10^{23} \times 6 .63 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}}$

$= 3.99 \times 10^{5} J / mole = 399 KJ / mole$

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2022

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