A) 594 nm
B) 640 nm
C) 700 nm
D) 494 nm
Correct Answer: D
Solution :
[d] Energy required for 1 \[C{{l}_{2}}\]molecule \[=\frac{242\times {{10}^{3}}}{{{N}_{A}}}J\] \[E=\frac{hc}{\lambda }\]or \[\lambda =\frac{hc}{E}\] \[=\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}\times 6.02\times {{10}^{23}}}{242\times {{10}^{3}}}\] \[=494\times {{10}^{-9}}m=494nm\]
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