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Murljashka [212]
3 years ago
8

At one time, it was thought that positive charge was spread throughout the atom. This was the plum-pudding model. Rutherford's e

xperiment to test this theory involved shooting positively charged particles toward the atom. Describe what the results of this experiment would have been if the plum-pudding model were correct.
Physics
2 answers:
laiz [17]3 years ago
7 0
The 2 positively charged particles would repel each other.
dezoksy [38]3 years ago
4 0

Answer:

1) All the alpha particles must have deviated at some angle from their initial path.

2) None of the alpha particle will go out undeviating

Explanation:

In Rutherford's experiment he bombarded alpha particles on the nucleus of gold atom and then he observed that most of the alpha particles moves out undeviating while few of them shows small deflection.

But the most shocking part of experiment was that he saw that one out of 10,000 alpha particles are such which retrace their initial path i.e. those alpha particles are deviated nearly 180 degree angle.

Now these all observations will concluded as

1. Most of the part of atom is vacant and hence atom is not made up of positively charged.

2. Electrons revolves around the nucleus in circular orbits

3. The positive charge of the nucleus is concentrated at the nucleus of of atom which is a very small and dense volume of atom.

So these results are totally opposite of plum pudding model because as per plum pudding model the atom is a positively charged sphere and electrons are randomly placed in it.

If this model is assumed to be correct then in that case the observations must be like following

1) All the alpha particles must have deviated at some angle from their initial path.

2) None of the alpha particle will go out undeviating

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The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
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v is the final speed of the electron
e is the electron charge
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Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
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where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
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Eddi Din [679]

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Answer:

The deformation is 0.088289 m

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Explanation:

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So, let us take the average value

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The deformation is 0.088289 m

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