Question
According to law of photochemical equivalence the energy absorbed ( in $$ergs/mole$$ ) is given as $$\left( {h = 6.62 \times {{10}^{ - 27}}ergs,\,c = 3 \times {{10}^{10}}cm\,\,{s^{ - 1}},{N_A} = 6.02 \times {{10}^{23}}mo{l^{ - 1}}} \right)$$
A.
$$\frac{{1.956 \times {{10}^{16}}}}{\lambda }$$
B.
$$\frac{{1.19 \times {{10}^8}}}{\lambda }$$
C.
$$\frac{{2.859 \times {{10}^5}}}{\lambda }$$
D.
$$\frac{{2.859 \times {{10}^{16}}}}{\lambda }$$
Answer :
$$\frac{{1.19 \times {{10}^8}}}{\lambda }$$
Solution :
$$\eqalign{
& E = \frac{{hc}}{\lambda } \times {N_A} \cr
& = \frac{{6.62 \times {{10}^{ - 27}} \times 3 \times {{10}^{10}} \times 6.02 \times {{10}^{23}}}}{\lambda } \cr
& = \frac{{1.19 \times {{10}^8}}}{\lambda } \cr} $$