One of the most worrisome waste products of a nuclear reactor is plutonium
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27,586
<h3>Further explanation</h3><u>Given:</u>
A single gold atom has a diameter of
From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately
<u>Question:</u>
How many atoms thick were Rutherford's foil?
<u>The Process:</u>
Convert thickness from mm to cm.
The number of atoms is calculated from gold foil thickness divided by the atomic diameter.
Therefore, we get an atomic thickness of 27,586 atoms.
<u>Notes:</u>
- In 1909-1910, Ernest Rutherford with two of his assistants, namely Hans Geiger and Ernest Marsden, conducted a series of experiments to find out more about the arrangement of atoms. They fired at a very thin gold plate with high-energy alpha particles.
- One of their observations is that a small portion of alpha particles are reflected. This greatly surprised Rutherford. The reflected alpha particle must have hit something very dense in the atom. This fact is incompatible with the atomic model proposed by J.J. Thomson where the atoms are described as homogeneous in all parts with electrons and protons evenly distributed.
- In 1911, Rutherford was able to explain the scattering of alpha rays by proposing ideas about atomic nuclei. According to him, most of the mass and positive charge of atoms are concentrated at the center of the atom, hereinafter referred to as the nucleus.
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Keywords: if a single gold atom, has a diameter of 2.9 x 10⁻⁸ cm, how many, atoms thick, Rutherford's foil, his scattering experiment
4.3 x 10⁻⁸dm⁻²
Explanation:
Problem:
To solve this problem, we have apply the right rule of indices operation.
If we have an indices in the form;
= (
)
Solving this problem:
= x 10
)
= 0.43 x 10⁻⁷
= 4.3 x 10⁻¹ x 10⁻⁷
= 4.3 x 10⁻⁸
for the units:
=
The solution is: 4.3 x 10⁻⁸dm⁻²
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