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 $ 1010 cm s$-$1, NA = 6.02 $ \times $ 10$-$23 mol$-$1)

Question

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 $ 10¹⁰ cm s^$-$1, N_A = 6.02 $ \times $ 10^$-$23 mol^$-$1)

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

${{2.859 \times {{10}^5}} \over \lambda }$

${{2.859 \times {{10}^{16}}} \over \lambda }$

${{1.196 \times {{10}^{16}}} \over \lambda }$

Solution

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