What caused rutherford to propose a revised model of the atom
1 answer:
Explanation:
Rutherford proposed a revised model for the atom, called the planetary model. The previous model of the atom was Thomson's Plum Pudding Model which consisted of freely moving positive and negative charges inside the atom.
Rutherford proposed his model after an experiment he conducted called the Gold Foil Experiment. This experiment consisted of a thin gold sheet into which alpha particles were shot upon and they were detected by a sensor. The image attached will give a better explanation of this. In this experiment he shot a beam of alpha particles(helium nucleus) at a thin sheet of gold. Rutherford hypothesised that there should be minimum deflection of the positively charged alpha particles occuring due to the repulsion of the alpha particle with the positive charges in the thin gold sheet. This was not the case.
However what he found was that most of alpha particles went straight through the thin sheet of gold but some were reflected back to him. This surprised him. Hence he proposed that most of the atom must be empty space as most of the alpha particles went straight through the sheet and there must be a heavy nucleus inside the atom causing the alpha particles to bounce back.
You might be interested in
Answer:
a) I= I₀ (cos²θ - cos⁴θ) b) 75.5º
Explanation:
a) For this exercise we must use Malus's law
I = I₀ cos² θ
where tea is the angle between the two polarizers.
We apply this expression to our case
* Polarizer 1 suppose that it is vertical and polarizer 2 (intermediate) is at an angle θ with respect to the vertical
I₁ = I₀ cos² θ
* We analyze for the polarity 2 and the last polarizer 3 which indicate that it must be at 90º from the first one, therefore it must be horizontal.
The angle of polarizers 2 and 3 is θ' measured from the horizontal, if we measure with respect to the vertical
θ₂ = 90- θ’ = θ
fiate that in the exercise we must take a reference system and measure everything with respect to this system.
I = I₁ cos² θ'
we substitute
I = (I₀ cos² tea) cos² (θ - 90)
cos (θ -90) = cos θ cos 90 + sin θ sin 90 = sin θ
I = Io cos² θ sin² θ
1= cos²θ+ sin²θ
sin²θ = 1 - cos²θ
I= I₀ (cos²θ - cos⁴θ)
b) to find when the intensity is maximum,
we can use that we have an extreme point when the drift is zero
= 0
\frac{dI}{d \theta}= Io (2 cos θ - 4 cos³θ) = 0
whereby
cos θ - 2 cos³ θ = 0
cos θ ( 1 - 2 cos² θ) = 0
The zeros of this function are in
θ = 90º
1-2cos²θ =0 cos θ = 0.25 θ = 75.5º
Let's analyze this two results for the angle of 90º the intnesidd is zero with respect to the first polarizer, so it is not an acceptable solution.
Consequently, the angle that allows the maximum intensity to pass is 75.5º