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}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 } $



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