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yan [13]
3 years ago
14

the majority of the alpha particles passed through with no deflection. what does this suggest about the structure of the atom?

Physics
2 answers:
Goryan [66]3 years ago
8 0
Rutherford's gold foil experiments (and other metal foil experiments) involved firing positively charged alpha particles at a piece of gold/metal foil. The alpha particles that were fired at the gold foil were positively charged. Most of the time, the alpha particles would pass through the foil without any change in their trajectories, which is what was expected if JJ Thomson's plum pudding model of the atom was correct. However, occasionally the alpha particles would be deflected to some degree, and sometimes an alpha particle would bounce back directly toward the experimenter. Rutherford likened this to firing a 15-inch artillery shell at a sheet of tissue paper and the shell came back to hit you.
In order for the alpha particles to be deflected, they would have to hit or come near to a positively charged particle in the atom. These experiments led Rutherford to describe the atom as containing mostly empty space, with a very small, dense, positively charged nucleus at the center, which contained most of the mass of the atom, with the electrons orbiting the nucleus.
kvv77 [185]3 years ago
3 0
This suggests that "most of the space within the atom is empty"

Hope this helps!
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Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

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Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

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a = Jenny's acceleration = 1.0 m/s²

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20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

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