Physics, 25.02.2020 19:01 kiaraphilman2956

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In Rutherford's gold foil experiment, the alpha particles that traveled through the gold foil hit the screen behind it in more than one spot. Explain the significance of this finding and the discovery it led to.

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Answer from: omar4538

In the gold foil experiment, Rutherford fired alpha particles (which are nuclei of helium) at a thing gold foil, and then he observed the angular distribution of the scattered alpha particles beyond the foil.

He made the following observations:

Most of the alpha particles passed through the foil mostly undeflectedSome of the alpha particles were instead deflected at very large anglesA few of the alpha particles were even reflected back to the original direction

According to these observations, he made the following conclusions:

The vast majority of the space in an atom is empty, as most of the the alpha particles pass undeflectedThe atoms have a central nucleus containing a large positive charge, to explain the large deflections of some alpha particlesThe nucleus of the atom is very tiny and contains most of the mass of the atom, to explain the alpha particles reflected back

This experiment therefore contribute to improve the model of the atom: the previous model was the Thomson model (plum-pudding model), but the experiment proved that this was not the right model, and the experiment lead to a new model, in which the atom is believed to consists of a tiny, positive nucleus surrounded by negatively charged electrons orbiting around the nucleus at a very large distance from it.

Answer from: Quest

true is the correct answer

hope i

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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