SATHEE: Atomic Structure Question 376

Atomic Structure Question 376

Question: According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as $(h = 6.62 \times 10^{- 27}$ ergs, $c = 3 \times 10^{10} c m s^{- 1}$ , $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ )

Options:

A) $\frac{1.956 \times 10^{16}}{λ}$

B) $\frac{1.19 \times 10^{8}}{λ}$

C) $\frac{2.859 \times 10^{5}}{λ}$

D) $\frac{2.859 \times 10^{16}}{λ}$

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Answer:

Correct Answer: B

Solution:

$E = \frac{h c}{λ} \times N_{A}$

$= \frac{6.62 \times 10^{- 27} \times 3 \times 10^{10} \times 6.02 \times 10^{23}}{λ}$

$= \frac{1.19 \times 10^{8}}{λ}$

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Atomic Structure Question 376

Question: According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as (h=6.62×10−27 ergs, c=3×1010cms−1 , NA=6.02×1023mol−1 )

Options:

Answer:

Solution:

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Question: According to law of photochemical equivalence the energy absorbed (in ergs/mole) is given as $(h = 6.62 \times 10^{- 27}$ ergs, $c = 3 \times 10^{10} c m s^{- 1}$ , $N_{A} = 6.02 \times 10^{23} m o l^{- 1}$ )