In famous Millikan's oil-drop method for determining the charge on an electron, the experimental measurements show that the oil-drops gain whole or integral number of charges, because.

The charges are multiples of the smallest charge

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The oil-drop is never stationary

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The oil-drop falls with a terminal velocity

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The oil drops are circular

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Solution

The correct option is A The charges are multiples of the smallest charge
In famous Millikan's oil-drop method, charges are multiples of smallest charge which show that oil-drops gain whole or integral no. of charges.

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Similar questions

Q. In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates.The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of

1.6 \times 10^{- 19} C

the charge of the electron. For this, he won the Nobel prize. Extra electrons on this particular oil drop (given the presently known charge of the electron) are :

1.6 \times 10^{- 19} C

the charge of the electron. For this, he won the Nobel prize. If a drop of mass

1.08 \times 10^{- 14} k g

remains stationary in an electric field of

1.68 \times 10^{5} N C^{- 1}

, then the charge of this drop is :

Assume that Millikan performed the oil drop experiment. In the oil drop experiment, the following charges(in arbitrary units) were found on a series of oil droplets.

The data obtained is as follows:

2 $\times$ $10^{- 19}$ C
4 $\times$ $10^{- 19}$ C
6 $\times$ $10^{- 19}$ C

Calculate the magnitude of the charge on the electron.

Q. In a Millikan's oil drop experiment, the charge on an oil drop is calculated to be

- 6.4 \times 10^{- 19} C

. The number of excess electrons on the drop is:

Q. An oil drop has 7.99

\times

10^{- 19}

C charge. The number of electrons that the oil drop has is .

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In famous Millikan's oil-drop method for determining the charge on an electron, the experimental measurements show that the oil-drops gain whole or integral number of charges, because.

The correct option is A The charges are multiples of the smallest chargeIn famous Millikan's oil-drop method, charges are multiples of smallest charge which show that oil-drops gain whole or integral no. of charges.

The correct option is A The charges are multiples of the smallest charge
In famous Millikan's oil-drop method, charges are multiples of smallest charge which show that oil-drops gain whole or integral no. of charges.