Question

According to law of photochemical equivalence the energy absorbed (in ergs / mole) is given as $\left(h =6.62\times10^{-27} ergs, c = 3 × 10^{10} cm s^{-1}, N_{A}=6.02× 10^{23}mol^{-1} \right)$

• A. $\frac{1.196\times10^{8}}{\lambda}$
• B. $\frac{2.859\times10^{5}}{\lambda}$
• C. $\frac{2.859\times10^{16}}{\lambda}$
• D. $\frac{1.196\times10^{16}}{\lambda}$

Answer 1

Answer: (A)

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