Atomic Structure Question 233
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}cm{s^{-1}} $ , $ N_{A}=6.02\times 10^{23}mo{l^{-1}} $ )
Options:
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 } $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ E=\frac{hc}{\lambda }\times N_{A} $
$ =\frac{6.62\times {10^{-27}}\times 3\times 10^{10}\times 6.02\times 10^{23}}{\lambda } $
$ =\frac{1.19\times 10^{8}}{\lambda } $
BETA
