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× 1023mol-1 )

Question

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× 1023mol-1 ) (A) (1.196×108/λ) (B) (2.859×105/λ) (C) (2.859×1016/λ) (D) (1.196×1016/λ)

Tardigrade Admin · Accepted Answer

According to stark Einstein's Law of photochemical equivalence, E=(h C/λ) × NA E=(6.62 × 10-27 × 3 × 1010 × 6.02 × 1023/λ) E=(1.196 × 108/λ)

Q. According to law of photochemical equivalence the energy absorbed (in ergs / mole) is given as $(h = 6.62 \times 1 0^{- 27} er g s, c = 3 \times 1 0^{10} c m s^{- 1}, N_{A} = 6.02 \times 1 0^{23} m o l^{- 1})$

$\frac{1.196 \times 1 0 ^{8}}{λ}$

$\frac{2.859 \times 1 0 ^{5}}{λ}$

$\frac{2.859 \times 1 0 ^{16}}{λ}$

$\frac{1.196 \times 1 0 ^{16}}{λ}$

According to stark Einstein's Law of photochemical equivalence,
$E = \frac{h C}{λ} \times N_{A}$
$E = \frac{6.62 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{10} \times 6.02 \times 1 0 ^{23}}{λ}$
$E = \frac{1.196 \times 1 0 ^{8}}{λ}$

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

λ1.196×108​

λ2.859×105​

λ2.859×1016​

λ1.196×1016​