A) 126 nm
B) 175 nm
C) 190 nm
D) 205 nm
Correct Answer: A
Solution :
BE of \[{{O}_{2}}\] per molecule\[=\frac{498\times {{10}^{3}}}{6.023\times {{10}^{23}}}J\] \[=826\times {{10}^{-19}}J\] Energy required for photochemical decom-position of one molecule of \[{{O}_{2}}=826\times {{10}^{-19}}J+2.5\times 1.6\times {{10}^{-19}}J\]\[=1.22\times {{10}^{-18}}J\] \[\lambda =\frac{hc}{E}=\frac{6.625\times {{10}^{-34}}\times 3\times {{10}^{8}}}{1.22\times {{10}^{-18}}}=126\,nm\]You need to login to perform this action.
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