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According to law of photochemical equivalence the energy absorbed (in erg/ mole $)$ is given as $\left(h=6.62 \times 10^{-27}\right.$ ergs, $\mathrm{c}=3 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}, \mathrm{~N}_{\mathrm{A}}=6.02 \times 10^{-23}$ $\left.\mathrm{mol}^{-1}\right)$ :
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Verified Answer
The correct answer is:
$\frac{1.196 \times 10^8}{\lambda}$
$$
\begin{aligned}
& \mathrm{E}=\frac{h c \mathrm{~N}_{\Lambda}}{\lambda} \\
& =\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda} \\
& =\frac{1.196 \times 10^8}{\lambda} \mathrm{ergs} / \mathrm{mol}
\end{aligned}
$$
\begin{aligned}
& \mathrm{E}=\frac{h c \mathrm{~N}_{\Lambda}}{\lambda} \\
& =\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda} \\
& =\frac{1.196 \times 10^8}{\lambda} \mathrm{ergs} / \mathrm{mol}
\end{aligned}
$$
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