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Chapter 27: Problem 85

If \(g_{E}\) and \(g_{m}\) are the accelerations due to gravity on the surfaces of the earth and moon respectively and if Millikan's oil drop experiment and the performed on the two surfaces one will be find the ratio \(=\frac{\text { electronic charge on the moon }}{\text { electronic charge on the earth }}\) to be (a) 1 (b) 0 (c) \(\frac{g_{E}}{g_{m}}\) (d) \(\frac{g_{m}}{g_{E}}\)

Short Answer

Expert verified
The ratio is 1, as electronic charge is the same everywhere.

Step by step solution

01

Understanding the Problem

We need to find the ratio of the electronic charge on the moon to the electronic charge on the earth from Millikan's oil drop experiment conducted on both surfaces. We have four choices.
02

Analyzing Millikan鈥檚 Experiment

In Millikan's oil drop experiment, the electronic charge measured, denoted by \( e \), is a fundamental constant. This means that \( e \) does not change regardless of the location, whether on Earth or the moon.
03

Applying the Concept of Fundamental Charge

Since the electronic charge \( e \) is constant at any point in the universe, the ratio of the electronic charge on the moon to the electronic charge on the earth is given by \( \frac{e_{\text{moon}}}{e_{\text{earth}}} = \frac{e}{e} = 1 \).
04

Conclusion: Final Answer

The ratio \( \frac{e_{\text{moon}}}{e_{\text{earth}}} \) is therefore 1. This means the electronic charge does not depend on the gravitational field where the experiment is conducted.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electronic Charge
Electronic charge is a fundamental property of particles, often associated with protons, electrons, and other subatomic particles. The basic unit of electronic charge, identified as \( e \), has a fixed magnitude of approximately \( 1.6 \times 10^{-19} \) coulombs for an electron. This constant remains unchanged irrespective of any external conditions, such as the gravitational field or location.
  • Electronic charge is a constant; it doesn鈥檛 vary from one location to another, such as between the Earth and the Moon.
  • In Millikan's oil drop experiment, electronic charge was measured to be the same on all Earth-like tests.
  • Because of its constant nature, the ratio of electronic charge on the Moon to that on Earth remains \( 1 \).
This means it doesn't matter where the experiment is conducted; on the Moon or Earth, the electronic charge will still be the same.
Gravity
Gravity is a natural force through which objects with mass attract each other. On Earth, the gravitational force attracts objects towards the center with an acceleration known as acceleration due to gravity. The strength of gravity can vary depending on the celestial body, such as Earth or the Moon.
  • Gravity is responsible for the weight of objects.
  • On Earth, gravity provides an acceleration of 鈮� \(9.81\,\text{m/s}^2\).
  • The gravitational forces are weaker on the Moon, causing objects to weigh less.
Understanding gravity is crucial for many experiments, including those related to space exploration and mechanics.
Acceleration Due to Gravity
Acceleration due to gravity refers to the rate at which objects accelerate towards the center of a celestial body because of the gravitational force. It is denoted by \( g \). It varies across different planets and celestial bodies. For instance, the Moon has a much lower gravitational pull than Earth.
  • On Earth, \( g_E = 9.81\,\text{m/s}^2\).
  • On the Moon, acceleration due to gravity is approximately one-sixth that of Earth's, known as \( g_m \) 鈮� \( 1.63\,\text{m/s}^2\).
  • The differences in acceleration create variations in weight measurements of the same object on different celestial bodies.
These differences are significant when calculating object behaviors in varying gravitational fields. However, the electronic charge remains unaffected by these gravitational changes.
Comparison of Gravitational Effects
Comparing the gravitational effects on different celestial bodies helps us understand how mass and gravity interact. When conducting experiments like Millikan's oil drop, this comparison becomes essential to identify what factors affect the outcomes and what remains consistent.
  • On Earth, gravity is stronger, influencing the fall speed of objects.
  • On the Moon, the weaker gravitational force results in slower falls.
  • Nonetheless, these differences in gravity do not influence the electronic charge value in any location.
In Millikan's experiment, even though gravity affects how the oil drops move, the electronic charge remains consistent across different gravitational fields. This illustrates the fundamental and unchanging nature of the electronic charge across various environments.

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Most popular questions from this chapter

The radioactivity of a given sample of whisky due to tritium (half-life \(12.3 \mathrm{yr}\) ) was found to be only \(3 \%\) of that measured in a recently purchased bottle marked 7 years old. The sample must have been prepared about (a) 220 years back (b) 300 years back (c) 400 years back (d) 70 years back

The normal activity of living carbon containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive \({ }_{6}^{14} \mathrm{C}\) present with the stable carbon isotope \({ }_{6}^{12} \mathrm{C}\). when the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life \((5730 \mathrm{yr})\) of \({ }_{6}^{14} \mathrm{C}\) and the measured activity, the age of the specimen can be approximately estimated. This is the principle of \({ }_{6}^{14} \mathrm{C}\) dating used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilization. (a) \(5224 \mathrm{yr}\) (b) \(4224 \mathrm{yr}\) (c) \(8264 \mathrm{yr}\) (d) \(6268 \mathrm{yr}\)

The binding energy of a H-atom, considering an electron moving around a fixed nuclei (proton), is \(B=\frac{m e^{4}}{8 n^{2}{ }_{0}^{2} h^{2}} \cdot(m=\) proton mass \()\) If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be \(B=-\frac{M e^{4}}{8 n^{2} \varepsilon_{0}^{2} h^{2}}(M=\) proton mass \() .\) This last expression is not correct because (a) \(n\) would not be integral (b) Bohr-quantisation applies only to electron (c) the frame in which the electron is at rest is not inertial. (d) the motion of the proton would not be in circular orbits, even approximately.

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A cathode emits \(1.8 \times 10^{14}\) electrons per second, when heated. When \(400 \mathrm{~V}\) is applied to anode all the emitted electrons reach the anode. The charge on electron is \(1.6 \times 10^{-19}\) C. One maximum anode current is (a) \(2.7 \mu \mathrm{A}\) (b) \(29 \mu \mathrm{A}\) (c) \(72 \mu \mathrm{A}\) (d) \(29 \mathrm{~mA}\)
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