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Milkman's oil drop experiment gives the ...

Milkman's oil drop experiment gives the value of

A

e

B

m

C

e/m

D

e-m

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### Step-by-Step Solution: 1. **Understanding the Experiment**: The Millikan oil drop experiment was designed to measure the charge of the electron. It involved a setup with two charged plates (one positive and one negative) and a small hole between them. 2. **Setup Description**: In the experiment, oil droplets were sprayed into the space between the plates. The droplets would fall under the influence of gravity, but when a voltage was applied across the plates, the electric field would exert a force on the charged oil droplets. 3. **Observation of Oil Drops**: The oil drops that passed through the small hole between the plates were observed using a microscope. Some of these drops would be attracted to the negatively charged plate, while others would remain suspended or fall due to gravity. 4. **Ionization of Oil Drops**: To determine the charge on the oil drops, Millikan used radiation to ionize the air, which allowed some of the oil droplets to gain a charge. This ionization was crucial for the experiment as it enabled the measurement of the charge on the droplets. 5. **Calculating Charge to Mass Ratio (E/m)**: By balancing the gravitational force acting on the oil drops with the electric force due to the electric field between the plates, Millikan was able to calculate the charge-to-mass ratio (E/m) of the oil droplets. This ratio is essential in determining the fundamental charge of the electron. 6. **Conclusion**: The main outcome of Millikan's oil drop experiment was the determination of the charge-to-mass ratio (E/m) of the electron, which was a significant contribution to the understanding of atomic structure. ### Final Answer: The Millikan oil drop experiment gives the value of the charge-to-mass ratio (E/m) of the electron. ---

### Step-by-Step Solution: 1. **Understanding the Experiment**: The Millikan oil drop experiment was designed to measure the charge of the electron. It involved a setup with two charged plates (one positive and one negative) and a small hole between them. 2. **Setup Description**: In the experiment, oil droplets were sprayed into the space between the plates. The droplets would fall under the influence of gravity, but when a voltage was applied across the plates, the electric field would exert a force on the charged oil droplets. 3. **Observation of Oil Drops**: The oil drops that passed through the small hole between the plates were observed using a microscope. Some of these drops would be attracted to the negatively charged plate, while others would remain suspended or fall due to gravity. ...
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In Millikan's oil drop experiment, the charge on a oil drop was found due to an experimental error, to be 8.88 xx 10^(-18)C . Why can't such a quantity of charge be present on the oil drop ? Explain. (Charge on an electron is 1.6 xx 10^(-19)C )

An oil drop of radius 2 mm with a density 3 g cm^(-3) is held stationary under a constant electric field 3.55 xx 10^5 V m^(-1) in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess ? Consider g=9.81 m//s^2

Knowledge Check

  • Millikan's oil drop experiment gives the value of

    A
    e
    B
    m
    C
    elm
    D
    em
  • In Millikan's oil drop experiment, we make use of:

    A
    Ohm's law
    B
    Ampere's law
    C
    Stoke's law
    D
    Faraday's law
  • In a Millikan's oil drop experiment the charge on an oil drop is calculated to be 16.35 xx 10^(-19) C . The number of excess electrons on the drop is

    A
    `3.9`
    B
    `4`
    C
    `4.2`
    D
    `6`
  • Similar Questions

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    An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 xx 10 ^(4) NC^(-1) in Millikan's oil drop experiment. The density of the oils is 1.26 g cm ^(-3) Estimate the radius of the drop (g= 9.81 m s ^(-2) and e = 1.60 xx 10 ^(-19) C )

    An oil drop of 12 excess electrons is held stationaty under a constant electric field of 2.55xx10^(4) NC^(-1) in Millikan's oil drop experi,ment. The density of the oil is 1.26 g cm^(-3) . Estimate the radius of the drop. (g = 9.81 ms^(-2) , e = 1.60xx10^(19) C)

    In Millikan's oil drop experiment, the distance between the metal plates, A and B to which an electric potential is applied such that A is positive and B is negative is 5 mm. An oil drop is found to be suspended at a distance of 2 mm from B. Predict the change in the position of the oil drop when there is a sudden drop or rise in potential. Justify.

    Millikan's oil drop experiment established that

    An oil drop of 10 excess electron is held stationary under a consatnt electric field of 3.6xx10^(4)NC^(-1) in Millikan's oil drop experiment. The density of oil is 1.26gcm^(-3) . Radius of the oil drop is (Take, g=9.8ms^(-2), e=1.6xx10^(-19)C )