A) 2 mA, towards left
B) 2 mA, towards right
C) 1 mA, towards right
D) 2 mA, towards left
Correct Answer: C
Solution :
Number of electrons \[s={{\left( \frac{{{K}_{sp}}}{4} \right)}^{1/3}}={{\left( \frac{32\times {{10}^{-12}}}{4} \right)}^{1/3}}=2\times {{10}^{-4}}M\] Number of protons \[k={{\wedge }_{eq}}.C\] Now, the current is given by the relation \[=\left( 91{{\Omega }^{-1}}c{{m}^{2}}e{{q}^{-1}} \right)\left( \frac{2.54}{159/2\times 1000}eq.c{{m}^{-3}} \right)\] \[=2.9\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}\] \[\frac{{{\left( {{t}_{1/2}} \right)}_{1}}}{{{\left( {{t}_{1/2}} \right)}_{2}}}={{\left[ \frac{{{a}_{2}}}{{{a}_{1}}} \right]}^{n-1}}\] \[\left( {{\text{t}}_{\text{1/2}}} \right)\text{=0}\text{.1s}\text{. }{{\text{a}}_{1}}=400\] Now, due to excess the charge on electrons, the direction of the current will be towards right
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