Chemistry, 01.09.2019 07:00 OliTepley8032

If a single gold atom has a diameter of 2.9×x10−8 cm, how many atoms thick was rutherford's foil

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Answer from: goeringwilliam5470

27,586

Further explanation

Given:

A single gold atom has a diameter of $\boxed{ \ 2.9 \times 10^{-8} \ cm. \ }$

From a reference, the Rutherford gold foil used in his scattering experiment had a thickness of approximately $\boxed{ \ 8 \times 10^{-3} \ mm. \ }$

Question:

How many atoms thick were Rutherford's foil?

The Process:

Convert thickness from mm to cm.

$\boxed{ \ 8 \times 10^{-3} \ mm = 8 \times 10^{-3} \times 10{-1} \ cm \ } \rightarrow \boxed{ \ 8 \times 10^{-4} \ cm \}$

The number of atoms is calculated from gold foil thickness divided by the atomic diameter.

$\boxed{ \ = \frac{8 \times 10^{-4} \ cm}{2.9 \times 10^{-8} \ cm} \ }$

$\boxed{ \ =2.7586 \times 10^4 \ atoms \ }$

Therefore, we get an atomic thickness of 27,586 atoms.

Notes:

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.Learn moreThe energy density of the stored energy The theoretical density of platinum which has the FCC crystal structure. Compound microscope

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

Answer from: preservations

Rutherford’s foil was $\boxed{{\text{27586 atoms}}}$ thick.

Further Explanation:

Rutherford’s experiment:

Alpha particles were used by Rutherford in his experiment for demonstrate the structure of atom. This experiment was put forward to overcome the drawbacks of Thomson’s atomic model.

Postulates of Rutherford’s experiment:

1.Atom’s positive charge is located in very small volume and this is called atomic nucleus. It is the central part of any atom.

2.Electrons are allowed to revolve around atomic nucleus in definite orbits.

3.There is strong electrostatic force of attraction between electrons and atomic nucleus.

Drawbacks of Rutherford’s experiment:

1.This model did not provide any description about the distribution of electrons in orbits.

2.Stability of atom was not explained.

3. It failed to explain Maxwell’s theory of electromagnetic radiation.

We know, thickness of gold foil that was used by Rutherford in his experiment was $8 \times {10^{ - 3}}{\text{ mm}}$ .

The thickness of gold foil is converted from mm to cm. The conversion factor for this is,

$1{\text{ mm}} = {10^{ - 1}}{\text{ cm}}$

Therefore thickness of gold foil can be calculated as follows:

$\begin{aligned}{\text{Thickness of gold foil}} &= \left( {8 \times {{10}^{ - 3}}{\text{ mm}}} \right)\left( {\frac{{{{10}^{ - 1}}{\text{ cm}}}}{{1{\text{ mm}}}}} \right) \\&= 8 \times {10^{ - 4}}{\text{ cm}} \\\end{aligned}$

Diameter of single gold atom is given as $2.9 \times {10^{ - 8}}{\text{ cm}}$ . Therefore number of gold atoms can be calculated as follows:

$\begin{aligned}{\text{Number of gold atoms}} &= \frac{{8 \times {{10}^{ - 4}}{\text{ cm}}}}{{2.9 \times {{10}^{ - 8}}{\text{ cm}}}} \\&= 27586.2{\text{ atoms}} \\& \approx {\text{27586 atoms}} \\\end{aligned}$

Hence Rutherford’s foil used in the gold-foil experiment was 27586 atoms thick, provided the diameter of single gold atom is $2.9 \times {10^{ - 8}}{\text{ cm}}$ .

Learn more:

The major contribution of Antoine Lavoisier to chemistry: Example of physical change:

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Atomic Structure

Keywords: Rutherford, atom, drawbacks, gold atoms, alpha particles, 27586 atoms, thick, Rutherford’s foil, diameter.

Answer from: Quest

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.

Answer from: Quest

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