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

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

give brainliest

water pollution kills aquatic life. aquatic life includes algae and other plant life that use photosynthesis. photosynthesis takes c02 out of the air.

c02is a greenhouse gas.

Answer from: Quest

explanation:

the largest database of organic compounds lists about 10 million substances, which include compounds originating from living organisms and those synthesized by chemists. the number of potential organic compounds has been estimated at 1060—an astronomically high number. the existence of so many organic molecules is a consequence of the ability of carbon atoms to form up to four strong bonds to other carbon atoms, resulting in chains and rings of many different sizes, shapes, and complexities. the simplest organic compounds contain only the elements carbon and hydrogen, and are called hydrocarbons. even though they are composed of only two types of atoms, there is a wide variety of hydrocarbons because they may consist of varying lengths of chains, branched chains, and rings of carbon atoms, or combinations of these structures. in addition, hydrocarbons may differ in the types of carbon-carbon bonds present in their molecules. many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory. we use hydrocarbons every day, mainly as fuels, such as natural gas, acetylene, propane, butane, and the principal components of gasoline, diesel fuel, and heating oil. the familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons. we can distinguish several types of hydrocarbons by differences in the bonding between carbon atoms. this leads to differences in geometries and in the hybridization of the carbon orbitals.

Another question on Chemistry

Chemistry, 21.06.2019 18:30

Asample of the male sex hormone testosterone, c19h28o2, contains 3.88×10^21 atoms of hydrogen.(a) how many atoms of carbon does it contain? (b) how many molecules of testosterone does it contain? (c) how many moles of testosterone does it contain? (d) what is the mass of this sample in grams?

Answers: 1

Answer

Chemistry, 21.06.2019 20:00

2h2s + 3o2 2so2 + 2h2o which option gives the correct mole ratios? h2s: so2 = 2: 2 and o2: h2o = 3: 2 h2s: so2 = 2: 3 and o2: h2o = 3: 2 h2s: so2 = 4: 4 and o2: h2o = 5: 4 h2s: so2 = 4: 6 and o2: h2o = 4: 4

Answers: 1

Answer

Chemistry, 23.06.2019 07:00

0.88 moles of n2o5 (g) was placed in a sealed 1.00 l vessel. calculate the equilibrium concentration of n2o5. no2, and o2 and the equilibrium constant after equilibrium has been reached by 65.0% of the n2o5 decomposing.

Answers: 1

Answer

Chemistry, 23.06.2019 07:30

Type the letter that represents the correct location for each particle type below. the neutron is found at __ the electron is found at __ the proton is found at __

Answers: 2

Answer

You know the right answer?

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

Questions