A) \[{{\log }_{e}}\frac{2}{5}\]
B) \[\frac{5}{{{\log }_{e}}2}\]
C) \[5{{\log }_{10}}2\]
D) \[5{{\log }_{e}}2\]
Correct Answer: D
Solution :
Fraction remains after n half-lives \[2NaOH+C{{O}_{2}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\] \[N{{a}_{2}}C{{O}_{3}}+2HCl\to 2NaCl+{{H}_{2}}O+C{{O}_{2}}\] \[{{p}^{5+}}.\] Taking log on both sides, we get \[8Al\to 8A{{l}^{3+}}+24{{e}^{-}}\] \[9F{{e}^{8/3+}}+24{{e}^{-}}\to 9Fe\] \[\underset{\left( \text{no lp} \right)}{\mathop{NH_{4}^{+}}}\,>\underset{\left( 1lp \right)}{\mathop{N{{H}_{3}}}}\,>\underset{\left( 2lp \right)}{\mathop{NH_{2}^{-}}}\,\] Now, let t? be the time after which activity reduces to half \[\frac{n{{'}_{He}}}{n{{'}_{C{{H}_{4}}}}}=\frac{1}{2}\sqrt{\frac{16}{4}}=\frac{1}{1}\] \[\frac{n{{'}_{He}}}{n{{'}_{S{{O}_{2}}}}}=\frac{1}{3}\sqrt{\frac{64}{4}}=\frac{4}{3}\]
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