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Millikan's Oil Drop Experiment

Millikan’s oi...

Question

Millikan’s oil drop experiment was conducted to measure the value of the:

A

e/m ratio

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B

Charge ‘e’ on an electron

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C

Mass of a proton

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D

Mass of an electron

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Solution

The correct option is B Charge ‘e’ on an electron
Millikan’s oil drop experiment was conducted to measure the charge on an electron. Since the value of e/m was known, the calculation of the value of charge ‘e’ lead to the calculation of the mass of an electron.

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Q. Millikan’s oil drop experiment was conducted to measure the value of the:

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 :

g_{E}

g_{M}

are the accelerations due to gravity on the surfaces of the earth and moon respectively and if Milikan's oil drop experiment could be performed on the two surfaces, one will find the ratio

\frac{electronic charge on the moon}{electronic charge on the earth}

to be:

(What Millikan did was to put a charge on a tiny drop of oil, and measure how strong an applied electric field had to be in order to stop the oil drop from falling.)

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. 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 :

Q. The oil drop experiment by R.A. Millikan was performed to find________.

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Millikan's Oil Drop Experiment

Standard XII Chemistry

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