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NEET · Chemistry · STD 11 - 2. structure of atom
According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as \((h = 6.62 \times 10^{-27} \ ergs,\)\( \ c = 3 \times cm s^{-1}, \)\(\ N_A = 6.02 \times 10^{23} \ mol^{-1})\)
- 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 \(\frac {1.196 \times 10^{16}}{\lambda}\)
Answer & Solution
Correct Answer
(A) \(\frac {1.196 \times 10^8}{\lambda}\)
Step-by-step Solution
Detailed explanation
\(E=\frac{h c N_{A}}{\lambda}\) \(=\frac{6.62 \times 10^{-27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{\lambda}\)
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