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15/08/2021 11:16 am
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According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h = 6.62 × 10-27 ergs, c = 3 × 1010 cm s-1, NA = 6.02 × 1023 mol-1)
(a) \(\frac{1.956 \times 10^{16}}{\lambda}\)
(b) \(\frac{1.19 \times 10^{8}}{\lambda}\)
(c) \(\frac{2.859 \times 10^{5}}{\lambda}\)
(d) \(\frac{2.859 \times 10^{16}}{\lambda}\)
1 Answer
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15/08/2021 11:20 am
Correct answer:(b) \(\frac{1.19 \times 10^{8}}{\lambda}\)
Explanation:
E = \(\frac{hc}{\lambda} \times N_A\)
= \(\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda}\)
\(\frac{1.19 \times 10^{8}}{\lambda}\)