A) \[2.48\times {{10}^{4}}nm\]
B) \[1.49\times {{10}^{3}}nm\]
C) \[2.48\times {{10}^{3}}nm\]
D) \[1.49\times {{10}^{4}}nm\]
Correct Answer: B
Solution :
In \[C{{H}_{4}},4\times B{{E}_{(C-H)}}=360kJ/mol\] \[\therefore B{{E}_{(C-H)}}=90\text{ }kJ/mol\] In \[{{C}_{2}}{{H}_{6}},B{{E}_{(C-C)}}+6\times B{{E}_{(c-H)}}=620kJ/mol\] \[\therefore B{{E}_{(C-C)}}=80KJ/mol\] \[\therefore B{{E}_{(C-C)}}=\frac{80\times {{10}^{3}}}{6.023\times {{10}^{23}}}J/molecule\] Now, \[E=\frac{hc}{\lambda }\] \[\therefore \lambda =\frac{6.626\times {{10}^{-34}}\times 3\times {{10}^{8}}\times 6.023\times {{10}^{23}}}{80\times {{10}^{3}}}\] \[\therefore \lambda =1.49\times {{10}^{3}}nm\]
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