A) \[4.68\times {{10}^{18}}\]
B) \[4.68\times {{10}^{15}}\]
C) \[4.68\times {{10}^{12}}\]
D) \[4.68\times {{10}^{9}}\]
Correct Answer: A
Solution :
Given, \[~i=25mA=0.0025\text{ }A\] \[t=\,\,60s\] \[Q=it\] \[=0.0025\times 60\] \[=1.5C\] number of electrons in 1.5 C \[=\frac{Q\times Avogadro\,number}{96500}\] \[=\frac{1.5\times 6.023\times {{10}^{23}}}{96500}\] \[=9.36\times {{10}^{18}}\] \[Ca\xrightarrow{{}}C{{a}^{2+}}+2{{e}^{-}}\] \[2{{e}^{-}}\] are required to deposit 1 Ca atom. \[\therefore \] number of Ca atoms deposited \[=\frac{no.\,of\,electrons}{2}\] \[=\frac{9.36\times {{10}^{18}}}{2}\] \[=4.68\times {{10}^{18}}\]
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