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

Answers: 1

Show answers

Answers

answered: 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

answered: 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.

answered: Guest

The answer to that would be magnesium

answered: Guest

There are no metaphors.

Other questions on the subject: Chemistry

Chemistry, 21.06.2019 22:30, colochaortiz20p7cajw

1.aluminum chloride (alcl3), and sodium hydroxide (naoh) can react to form aluminum hydroxide (al(oh)3) and sodium chloride (nacl). you have 13.4 g of aluminum chloride and 10.0 g of sodium hydroxide. answer the following questions: •what is the balanced equation for this reaction? •if you use all 13.4 g of aluminum chloride, how many grams of aluminum hydroxide can be formed? work must be shown to earn credit •if you use all 10.0 g of sodium hydroxide, how many grams of aluminum hydroxide can be formed? work must be shown to earn credit •how many grams of aluminum hydroxide will actually be made? which reagent is limiting? explain your answer.

Answers: 1

continue

Chemistry, 22.06.2019 13:00, jaylanmahone223

6. using 3 – 4 sentences explain (in your own words) why water expands when it freezes? 7. using your knowledge of colligative properties explain whether sodium chloride or calcium chloride would be a more effective substance to melt the ice on a slick sidewalk. use 3 – 4 sentences in your explanation.

Answers: 1

continue

Chemistry, 22.06.2019 17:30, sheazy3709

The polymer used for the nonstick surface of cooking utensils is 24.0%c and 76%f by mass. what is the empirical formula of this polymer?

Answers: 2

continue

Chemistry, 22.06.2019 18:50, cj31150631

Question 3(multiple choice worth 4 points) (04.04 lc) what does it mean when an element is reduced? it empties a valance shell, reducing its atomic radius. it gains electrons, reducing its overall charge. it increases electronegativity, reducing its ability to bond. it loses electrons, reducing its electron number.

Answers: 1

continue

Do you know the correct answer?

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