A) \[\text{1}{{0}^{-\text{19}}}\]
B) \[\text{1}{{0}^{-17}}\]
C) \[\text{1}{{0}^{-18}}\]
D) \[\text{1}{{0}^{17}}\]
Correct Answer: B
Solution :
We know that current (Z) = charge (q) \[\times \] time ... (i) Also, time \[(t)=\frac{1}{frequency\text{ }\left( n \right)}\] ... (ii) From Eqs. (i) and (ii), we get \[n=\frac{q}{I}\] Given, \[q=1.6\times {{10}^{-19}}C\], \[I=16\,mA=16\times {{10}^{-3}}A\] \[n=\frac{1.6\times {{10}^{-19}}}{16\times {{10}^{-3}}}={{10}^{-17}}\]
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