Structure of Atom - Result Question 2
####2. According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as $(h=6.62 \times 10^{-27}.$ ergs, $c=3 \times 10^{10} cm s^{-1}$, $N _{\text{A }}=6.02 \times 10^{23} mol^{-1}$ )
[NEET Kar. 2013]
(a) $\frac{1.196 \times 10^{16}}{\lambda}$
(b) $\frac{1.196 \times 10^{8}}{\lambda}$
(c) $\frac{2.859 \times 10^{5}}{\lambda}$
(d) $\frac{2.859 \times 10^{16}}{\lambda}$
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Solution:
- (b) $E=\frac{h c}{\lambda} \times N_A$
$=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda}$
$=\frac{1.19 \times 10^{8}}{\lambda} ergs mol^{-1}$
Law of photochemical equivalence is the fundamental principle relating to chemical reactions induced by light which states that for every mole of a substance that reacts $6.022 \times 10^{23}$ quanta of light are absorbed. The photochemical equivalence law is also sometimes called the stark-Einstein law.
