A) \[{{C}_{2}}{{H}_{5}}OH+HCl\xrightarrow{ZnC{{l}_{2}}}{{C}_{2}}{{H}_{5}}Cl+{{H}_{2}}O\]
B) \[{{C}_{6}}{{H}_{5}}Cl+C{{H}_{3}}COCl\xrightarrow{AlC{{l}_{3}}}{{C}_{6}}{{H}_{5}}COCl+C{{l}_{2}}\]
C) \[{{C}_{6}}{{H}_{5}}Br+Mg\xrightarrow{Ether}{{C}_{5}}{{H}_{5}}MgBr\]
D) \[FeSi{{O}_{3}}\]
Correct Answer: C
Solution :
Let \[IE={{E}_{\infty }}-{{E}_{1}}=0-{{E}_{1}}\] \[=2.18\times {{10}^{-18}}J\,at{{m}^{-1}}\] On comparing both sides, we get \[{{E}_{n}}=\frac{-2.18\times {{10}^{-18}}}{{{n}^{2}}}J\,at{{m}^{-1}}\] \[\Delta E={{E}_{4}}-{{E}_{1}}=-2.18\times {{10}^{-18}}\left( \frac{1}{{{4}^{2}}}-\frac{1}{{{1}^{2}}} \right)\] \[=2.044\times {{1}^{-18}}J\,at{{m}^{-1}}or\Delta E=hv\] \[v=\frac{\Delta E}{h}=\frac{2.044\times {{10}^{-18}}J}{6.625\times {{10}^{-34}}Js}\] unit of mass is \[=3.085\times {{10}^{15}}{{s}^{-1}}\]
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