(A) The energy of one mole of photons is given by the formula $E = \frac{h c N_A}{\lambda}$.Substituting the given values:$E = \frac{(6.62 \times 10^{

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$\frac{1.196 \times 10^8}{\lambda}$

(A) The energy of one mole of photons is given by the formula $E = \frac{h c N_A}{\lambda}$.Substituting the given values:$E = \frac{(6.62 \times 10^{-27} \ erg \ s) \times (3 \times 10^{10} \ cm \ s^{-1}) \times (6.02 \times 10^{23} \ mol^{-1})}{\lambda}$$E = \frac{11.95572 \times 10^6}{\lambda} \ erg \ mol^{-1}$$E = \frac{1.196 \times 10^8}{\lambda} \ erg \ mol^{-1}$.

Class 12 Chemistry Chemical Kinetics Photochemical reaction

Medium

According to the law of photochemical equivalence,the energy absorbed (in $ergs \ mol^{-1}$) is given as ($h = 6.62 \times 10^{-27} \ erg \ s$,$c = 3 \times 10^{10} \ cm \ s^{-1}$,$N_A = 6.02 \times 10^{23} \ mol^{-1}$):

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A
$\frac{1.196 \times 10^8}{\lambda}$
B
$\frac{2.859 \times 10^5}{\lambda}$
C
$\frac{2.859 \times 10^{16}}{\lambda}$
D
$\frac{1.196 \times 10^{16}}{\lambda}$

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Chemical Kinetics

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Photochemical reaction

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According to the law of photochemical equivalence,the energy absorbed (in $ergs \ mol^{-1}$) is given as ($h = 6.62 \times 10^{-27} \ erg \ s$,$c = 3 \times 10^{10} \ cm \ s^{-1}$,$N_A = 6.02 \times 10^{23} \ mol^{-1}$):

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