A) Zero, equal to the magnitude of charge on \[\alpha -\]particle
B) \[2e,\ 1.6\times {{10}^{-18}}C,\]
C) \[1.6\times {{10}^{-19}}C,\ 2.5e\]
D) \[1.5e,\ e\] (Here e is the electronic charge)
Correct Answer: B
Solution :
In Millikan?s experiment, the charges present on the oil drops are the integral multiples, so 2e and \[10e(1.6\times {{10}^{-18}}C)\] charges are present.
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