Atomic Models Alıştırma Soruları — JAMB UTME

Match each atomic model with its key defining feature or experimental foundation.

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Cevap

Thomson Model matches with electrons embedded in a positive sphere; Rutherford Model matches with the dense positive nucleus discovered via alpha particle scattering; Bohr Model matches with quantized non-radiating energy levels; Quantum Wave Model matches with standing matter waves in electron orbits.

Each atomic model corresponds directly to its historic milestone: Thomson proposed electrons scattered inside a uniform sphere of positive charge; Rutherford discovered the compact positive nucleus through alpha scattering experiments; Bohr introduced quantized stationary states to account for emission spectral lines; and the Quantum Wave Model integrated de Broglie matter wave harmonics into orbital mechanics.

Adım Adım Çözüm

1

Identify Thomson's Plum Pudding Model features

Describes electrons distributed inside a broad, uniform cloud/sphere of positive charge.

Formulated after J.J. Thomson discovered the electron before the nucleus was known.

2

Identify Rutherford's Nuclear Model features

Demonstrated that positive charge is concentrated in a tiny nucleus because alpha particles bounced back at large angles.

Geiger-Marsden alpha scattering experiment disproved the uniform charge distribution model.

3

Identify Bohr's Model features

Postulated quantized non-radiating stationary orbits where angular momentum is restricted to integral multiples of

h / 2\pi

.

Successfully explained the discrete wavelengths of the hydrogen emission spectrum.

4

Identify the Quantum Wave Model features

Explains quantization by treating orbiting electrons as de Broglie standing waves around the nucleus.

Orbits are stable only when an integral number of electron wavelengths fit along the orbital circumference.

Anahtar Kavram

Characteristics and experimental foundations of atomic models

Soru 2Soru →

In a hydrogen atom modeled according to Bohr's postulates, an electron undergoes a transition from an initial excited state $n$ to the ground state $n=1$ , emitting a photon whose wavelength is $\lambda = \frac{16}{15R}$ , where $R$ is the Rydberg constant. What is the orbital angular momentum of the electron in its initial excited state prior to the transition?

\frac{h}{2\pi}

\frac{h}{\pi}

\frac{2h}{\pi}

\frac{4h}{\pi}

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Cevap: $\frac{2h}{\pi}$

Cevap

\frac{2h}{\pi}

Using the Rydberg formula

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)

with

\lambda = \frac{16}{15R}

, we find

\frac{15}{16} = 1 - \frac{1}{n^2}

, which gives

n = 4

. By Bohr's angular momentum quantization condition,

L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}

.

Adım Adım Çözüm

1

Apply the Rydberg formula for hydrogen to determine the principal quantum number

n

of the initial state.

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \implies \frac{15R}{16} = R \left( 1 - \frac{1}{n^2} \right) \implies 1 - \frac{1}{n^2} = \frac{15}{16} \implies \frac{1}{n^2} = \frac{1}{16} \implies n = 4

.

The Rydberg formula links the photon wavelength emitted during a transition between energy levels to the principal quantum numbers.

2

Calculate the orbital angular momentum

L

for

n=4

using Bohr's quantization condition.

L = \frac{n h}{2\pi} = \frac{4 h}{2\pi} = \frac{2 h}{\pi}

.

Bohr's quantization postulate dictates that orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Anahtar Kavram

Bohr's Atomic Model: Spectral transitions and angular momentum quantization

Tahmini Süre:2m 0s

Soru 3Soru →

During Rutherford's alpha particle scattering experiment, the vast majority of alpha particles passed straight through the gold foil with little to no deflection. What conclusion about atomic structure was drawn directly from this specific observation?

The atom consists mostly of empty space

The positive charge is uniformly distributed throughout the atom like a sphere

Electrons orbit the nucleus in fixed, quantized energy levels

Alpha particles are absorbed by electrons, causing them to jump to higher energy states

Cevabı ve açıklamayı göster

Cevap: The atom consists mostly of empty space

Cevap

The atom consists mostly of empty space

In Rutherford's alpha particle scattering experiment, the fact that over 99% of alpha particles passed straight through the foil undeflected indicates that they encountered no massive positive obstacle. This directly proved that the volume of an atom consists predominantly of empty space, with positive charge concentrated in a tiny central region called the nucleus.

Adım Adım Çözüm

1

Analyze the experimental observation from Rutherford's experiment

The vast majority of positively charged alpha particles suffered no deflection when passing through gold foil.

Deflection occurs when alpha particles pass close to a dense positive charge.

2

Deduce the physical implication for the structure of the atom

Since almost no deflection occurred for most particles, they encountered no heavy concentration of positive charge or mass in their path.

This proves that the dense positively charged nucleus occupies an extremely small fraction of the total atomic volume, leaving the rest as empty space.

Anahtar Kavram

Rutherford's Alpha Scattering Experiment and Nuclear Atom Model

Soru 4Soru →

Match each historical atomic model on the left with its defining postulate, experimental outcome, or theoretical limitation on the right.

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Thomson's model matches the diffuse positive sphere disproved by alpha particle backscattering; Rutherford's model matches the dense nucleus with classical radiation collapse limitations; Bohr's model matches quantized angular momentum in non-radiating orbits; Sommerfeld's extension matches elliptical sub-shells and relativistic adjustments for fine-structure splitting.

Each model directly maps to its defining theoretical contribution or failure mechanism: Thomson's diffuse charge sphere failed under

\alpha

-particle scattering; Rutherford's nuclear atom suffered from classical radiation instability; Bohr's model introduced angular momentum quantization

L = n\hbar

; and Sommerfeld's model extended orbits to ellipses with relativistic velocity corrections to account for fine structure.

Adım Adım Çözüm

1

Analyze Thomson's Plum Pudding Model

Thomson proposed electrons embedded in a sea of positive charge. This continuous distribution could not account for

\alpha

-particles rebounding at angles greater than

90^\circ

.

Identify the historical assumption and experimental contradiction for Thomson's model.

2

Analyze Rutherford's Planetary Model

Rutherford deduced a concentrated positive core (nucleus). However, according to Maxwellian electrodynamics, orbiting electrons accelerate continuously, radiating energy until collapsing into the nucleus.

Identify the primary theoretical failure of classical planetary electron orbits.

3

Analyze Bohr's Quantized Model

Bohr introduced the non-classical postulate that electrons exist in stable stationary states with angular momentum

L = \frac{nh}{2\pi}

, accurately yielding the Rydberg formula for hydrogen.

Recognize the quantum postulate resolving Rutherford's radiation collapse.

4

Analyze Sommerfeld's Extension

To explain fine spectral line splitting not accounted for by circular Bohr orbits, Sommerfeld introduced elliptical orbits with azimuthal quantum numbers and relativistic mass variation at high electron velocities.

Connect fine-structure spectral features to relativistic elliptical orbital modifications.

Anahtar Kavram

Development and Limitations of Historical Atomic Models

Soru 5Soru →

In a hydrogen atom modeled according to Bohr's theory, an electron undergoes a transition from an excited state with an energy of $-1.51\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . Calculate the energy of the emitted photon in electron-volts ( $\text{eV}$ ).

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Cevap: 1.89

Cevap

The energy of the emitted photon is

1.89\text{ eV}

.

According to Bohr's atomic model, when an electron drops from an initial higher energy level

E_i

to a final lower energy level

E_f

, a photon is emitted carrying energy

E = E_i - E_f

. Substituting the given values gives

E = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

.

Adım Adım Çözüm

1

Identify the initial and final energy states of the electron.

E_i = -1.51\text{ eV}

and

E_f = -3.40\text{ eV}

.

The electron moves from a higher (less negative) energy state to a lower (more negative) energy state.

2

Apply Bohr's energy quantization formula for photon emission

E_{\text{photon}} = E_i - E_f

.

E_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}

.

By energy conservation, the energy lost by the transitioning electron equals the energy of the emitted photon.

Anahtar Kavram

Bohr's Energy Transition Postulate

Tahmini Süre:1m 0s

Soru 6Soru →

In J. J. Thomson's plum pudding model of the atom, how is the positive charge distributed throughout the atom?

It is uniformly spread throughout a spherical volume containing embedded electrons.

It is concentrated in a tiny, dense central core called the nucleus.

It orbits around a central cluster of negative charges in fixed energy levels.

It is localized entirely on the outer surface of a hollow atomic shell.

Cevabı ve açıklamayı göster

Cevap: It is uniformly spread throughout a spherical volume containing embedded electrons.

Cevap

The positive charge is uniformly spread throughout a spherical volume containing embedded electrons.

In J. J. Thomson's plum pudding model, the atom is envisioned as a continuous sphere of positive charge within which negatively charged electrons are embedded evenly to maintain overall electrical neutrality.

Adım Adım Çözüm

1

Recall the fundamental proposal of Thomson's atomic model (1897).

J. J. Thomson envisioned the atom as a sphere of positive electrification.

This accounted for atomic neutrality following his discovery of the electron.

2

Identify the arrangement of electrons within this positive sphere.

Electrons were thought to be embedded throughout the positive sphere like plums in a pudding.

This balanced the positive charge uniformly across the atomic volume.

Anahtar Kavram

Thomson's Plum Pudding Atomic Model

Tahmini Süre:45s

Soru 7Soru →

In Bohr's model of the hydrogen atom, the radius of a stationary orbit is proportional to $n^2$ , where $n$ is the principal quantum number. If the radius of the ground-state orbit ( $n = 1$ ) is $r_1$ , what is the radius of the orbit corresponding to the second excited state?

3r_1

4r_1

6r_1

9r_1

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Cevap: $9r_1$

Cevap

The radius of the electron's orbit in the second excited state is

9r_1

.

In Bohr's atomic model, the ground state corresponds to the quantum number

n = 1

. The excited states are numbered sequentially above the ground state: the first excited state is

n = 2

and the second excited state is

n = 3

. Since the orbital radius scales with the square of the principal quantum number (

r_n = n^2 r_1

), substituting

n = 3

yields

r_3 = 3^2 r_1 = 9r_1

. Thus, the option stating

9r_1

is correct.

Adım Adım Çözüm

1

Identify the principal quantum number

n

for the second excited state.

n = 3

The ground state corresponds to

n = 1

, the first excited state corresponds to

n = 2

, and the second excited state corresponds to

n = 3

.

2

Apply Bohr's radius formula for hydrogen-like atoms.

r_n = n^2 r_1

According to Bohr's postulates, the orbital radius is proportional to the square of the principal quantum number

n

.

3

Calculate

r_3

by substituting

n = 3

into the radius expression.

r_3 = 3^2 r_1 = 9r_1

Squaring

n = 3

gives

9

, making the radius 9 times the ground-state radius.

Anahtar Kavram

Bohr's quantization of orbital radius (

r_n \propto n^2

)

Tahmini Süre:1m 0s

Soru 8Soru →

Historical developments in atomic physics led to several distinct models of atomic structure. Match each atomic model on the left with its defining structural feature or experimental basis on the right.

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Cevap

Thomson's Model matches with embedded electrons in a positive sphere; Rutherford's Model matches with a dense positive nucleus from alpha scattering; Bohr's Model matches with quantized non-radiating orbits; Quantum Mechanical Model matches with electron probability orbitals.

Each historical atomic model is correctly matched to its fundamental feature: Thomson proposed electrons suspended in positive mass; Rutherford used alpha scattering to discover the compact nucleus; Bohr quantized electron orbits to explain spectral lines; and the Quantum Mechanical Model represents electrons via three-dimensional probability orbitals.

Adım Adım Çözüm

1

Identify Thomson's contribution

Thomson proposed the 'plum pudding' model where negative electrons are embedded inside a uniform positive charge sphere.

This preceded the discovery of the atomic nucleus.

2

Identify Rutherford's contribution

Rutherford discovered the central positive nucleus through the alpha particle deflection experiment.

Large deflections meant most atomic mass and positive charge concentrated at a tiny core.

3

Identify Bohr's contribution

Bohr added quantum conditions to planetary orbits so electrons remain stable without continuously radiating energy.

Quantized angular momentum explains discrete emission line spectra.

4

Identify the Quantum Mechanical Model contribution

Modern quantum mechanics replaces fixed circular orbits with wave functions and 3D probability clouds (orbitals).

Heisenberg's uncertainty principle rules out precise circular orbits.

Anahtar Kavram

Evolution of Atomic Models

Soru 9Soru →

According to Bohr's atomic model of the hydrogen atom, what condition must be satisfied by an electron moving in a stable stationary orbit?

Its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Its total energy is proportional to light intensity.

Its orbital frequency increases continuously as it radiates energy in the orbit.

Its total mechanical energy is zero in all allowed stationary orbits.

Cevabı ve açıklamayı göster

Cevap: Its orbital angular momentum is an integral multiple of $\frac{h}{2\pi}$ .

Cevap

An electron moves in a stable stationary orbit when its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Bohr's fundamental postulate states that an electron can revolve around the nucleus only in certain non-radiating orbits (stationary states) where its orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Adım Adım Çözüm

1

Recall Bohr's postulates for the hydrogen atom

Identify that stable orbits require quantization of orbital angular momentum.

Bohr introduced quantization to explain why orbiting electrons do not continuously emit radiation and collapse into the nucleus.

2

State the mathematical formula for angular momentum quantization

L = mvr = \frac{nh}{2\pi}

, where

n

is an integer (

1, 2, 3, \dots

) and

h

is Planck's constant.

Only specific discreet orbits meeting this condition are allowed stationary states.

Anahtar Kavram

Quantization of Angular Momentum in Bohr's Model

Tahmini Süre:45s

Soru 10Soru →

In Bohr's atomic model of the hydrogen atom, the radius of the ground state orbit ( $n = 1$ ) is $0.053\text{ nm}$ . According to de Broglie's condition for stationary electron orbits, what is the de Broglie wavelength of the electron in its second excited state? Express your answer in nanometers ( $\text{nm}$ ).

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Cevap: 1

Cevap

The de Broglie wavelength of the electron in its second excited state is

1.00\text{ nm}

.

The second excited state corresponds to the quantum number

n = 3

. According to Bohr's atomic model, the radius of the

n

-th orbit is given by

r_n = n^2 r_1

, yielding

r_3 = 3^2 \times 0.053\text{ nm} = 0.477\text{ nm}

. De Broglie explained Bohr's angular momentum quantization by showing that an integral number of electron matter-waves must fit around the orbital circumference:

2\pi r_n = n \lambda_n

. Solving for

\lambda_3

gives

\lambda_3 = \frac{2\pi r_3}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 1.00\text{ nm}

.

Adım Adım Çözüm

1

Identify the principal quantum number for the specified energy state

n = 3

The ground state corresponds to

n = 1

, the first excited state to

n = 2

, and the second excited state to

n = 3

.

2

Calculate the radius of the third stationary orbit

r_3 = 0.477\text{ nm}

In Bohr's model, the radius of the

n

-th orbit is proportional to

n^2

, so

r_3 = 3^2 \times 0.053\text{ nm} = 9 \times 0.053\text{ nm} = 0.477\text{ nm}

.

3

Apply de Broglie's standing wave condition for stationary orbits

\lambda_3 = \frac{2\pi r_3}{3}

De Broglie postulated that a stationary orbit contains an integral number of electron de Broglie wavelengths around its circumference:

2\pi r_n = n \lambda_n

.

4

Substitute values to compute the wavelength

\lambda_3 = 1.00\text{ nm}

\lambda_3 = \frac{2 \times 3.1416 \times 0.477\text{ nm}}{3} = 2\pi \times 3 \times 0.053\text{ nm} \approx 0.9992\text{ nm} \approx 1.00\text{ nm}

.

Anahtar Kavram

De Broglie Standing Wave Quantization in Bohr Atomic Model

Soru 11Soru →

During a head-on collision in Rutherford's $\alpha$ -particle scattering experiment, an $\alpha$ -particle of initial speed $v$ approaches a stationary heavy nucleus. The distance of closest approach achieved by the $\alpha$ -particle is $r_0$ . If the initial speed of the $\alpha$ -particle is increased to $2v$ , what will be the new distance of closest approach in terms of $r_0$ ?

\frac{r_0}{4}

\frac{r_0}{2}

2r_0

4r_0

Cevabı ve açıklamayı göster

Cevap: $\frac{r_0}{4}$

Cevap

The new distance of closest approach will be

\frac{r_0}{4}

.

At the distance of closest approach, the entire initial kinetic energy of the

\alpha

-particle is converted into electric potential energy:

E_k = \frac{1}{2}mv^2 = \frac{k_e q_1 q_2}{r}

. Rearranging for

r

gives

r = \frac{2k_e q_1 q_2}{m v^2}

, showing that

r

is inversely proportional to

v^2

. When the initial speed is doubled to

2v

, the kinetic energy increases by a factor of

2^2 = 4

. Consequently, the distance of closest approach is reduced to one-fourth of its initial value,

\frac{r_0}{4}

.

Adım Adım Çözüm

1

Apply conservation of mechanical energy at the distance of closest approach.

Initial kinetic energy equals electrostatic potential energy at distance

r_0

:

\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r_0}

.

At the distance of closest approach, the

\alpha

-particle momentarily stops, converting all kinetic energy into electric potential energy.

2

Express the distance of closest approach

r_0

in terms of initial speed

v

.

r_0 = \frac{2 q_1 q_2}{4\pi\varepsilon_0 m v^2} \propto \frac{1}{v^2}

.

Rearranging the energy conservation equation demonstrates that distance of closest approach is inversely proportional to

v^2

.

3

Substitute the new speed

v' = 2v

into the proportion.

r' \propto \frac{1}{(2v)^2} = \frac{1}{4v^2} = \frac{r_0}{4}

.

Doubling the speed quadruples the kinetic energy, reducing the distance required to bring the particle to rest by a factor of 4.

Anahtar Kavram

Distance of Closest Approach in Rutherford Scattering

Tahmini Süre:2m 0s

Soru 12Soru →

According to Bohr's model of the hydrogen atom, the total energy of an electron in the first excited state ( $n = 2$ ) is $-3.40\text{ eV}$ . Calculate the electric potential energy of the electron in this state, in electron-volts ( $\text{eV}$ ).

Cevabı ve açıklamayı göster

Cevap: -6.8

Cevap

The electric potential energy of the electron in the first excited state is

-6.80\text{ eV}

.

In Bohr's model of the hydrogen atom, an electron held in a circular orbit by electrostatic attraction has a kinetic energy

K = -E

and an electric potential energy

U = 2E

. Given a total energy

E = -3.40\text{ eV}

, multiplying by 2 yields

U = -6.80\text{ eV}

.

Adım Adım Çözüm

1

Identify the relationship between total energy and electric potential energy in the Bohr atomic model.

U = 2E

, where

U

is potential energy and

E

is total energy.

For an inverse-square electrostatic force binding an electron in a circular orbit, the virial theorem dictates that kinetic energy

K = -E

and potential energy

U = 2E

.

2

Substitute the given value of total energy

E = -3.40\text{ eV}

to solve for

U

.

U = 2 \times (-3.40\text{ eV}) = -6.80\text{ eV}

.

Direct multiplication yields the exact electric potential energy of the bound electron.

Anahtar Kavram

Energy components (kinetic, potential, and total energy) of an electron in Bohr's atomic model

Soru 13Soru →

In Rutherford's $\alpha$ -particle scattering experiment, most of the $\alpha$ -particles passed straight through the gold foil with negligible deflection, while a very small fraction was scattered through angles greater than $90^\circ$ . Which fundamental deduction about atomic structure was directly established by these rare, large-angle scatterings?

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Electrons revolve around the positive center in discrete, non-radiating stationary orbits.

The positive charge is distributed uniformly throughout a sphere of atomic dimensions containing embedded electrons.

Orbiting electrons continuously radiate energy, causing them to gradually spiral inward toward the center.

Cevabı ve açıklamayı göster

Cevap: The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Cevap

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Large-angle deflection (

>90^\circ

) requires an immense repulsive Coulomb force, which can only occur if the entire positive charge and virtually all atomic mass are concentrated in a tiny central region (the nucleus). If positive charge were spread out over the entire atomic volume, the maximum electric field would be far too weak to turn back high-velocity alpha particles.

Adım Adım Çözüm

1

Analyze the experimental observations from Rutherford's alpha-scattering experiment.

Most

\alpha

-particles pass undeflected (indicating mostly empty space), while a tiny fraction scatter at angles

>90^\circ

.

Understanding the physical cause of large-angle electrostatic deflection.

2

Relate electrostatic repulsive force to mass and charge distribution.

To turn around a fast-moving positive

\alpha

-particle (

He^{2+}

), it must experience a intense Coulomb repulsion

F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}

at very small distance

r

.

A diffuse charge distribution (Thomson model) yields weak electric fields that cannot cause wide-angle scattering.

3

Identify the structural conclusion drawn by Rutherford.

The entire positive charge and mass must reside in a massive, concentrated center (the nucleus).

Directly matches the core physical takeaway of the nuclear atomic model.

Anahtar Kavram

Rutherford Nuclear Model and Alpha Scattering Deduction

Soru 14Soru →

In Bohr's model of the hydrogen atom, the energy of an electron in a stationary orbit with principal quantum number $n$ is given by $E_n = -\frac{13.6}{n^2}\text{ eV}$ . What is the frequency of the photon emitted when an electron transitions from the $n = 4$ energy state to the $n = 2$ energy state? ( $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

6.15 \times 10^{14}\text{ Hz}

3.85 \times 10^{33}\text{ Hz}

2.05 \times 10^{14}\text{ Hz}

6.20 \times 10^{14}\text{ Hz}

Cevabı ve açıklamayı göster

Cevap: $6.15 \times 10^{14}\text{ Hz}$

Cevap

6.15 \times 10^{14}\text{ Hz}

The energy of the emitted photon corresponds to the difference between the initial energy level (

n = 4

) and the final energy level (

n = 2

), giving

\Delta E = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

. Converting

2.55\text{ eV}

to Joules yields

4.08 \times 10^{-19}\text{ J}

. Dividing this energy by Planck's constant (

6.63 \times 10^{-34}\text{ J}\cdot\text{s}

) produces a photon frequency of

6.15 \times 10^{14}\text{ Hz}

.

Adım Adım Çözüm

1

Calculate the energy levels for

n = 4

and

n = 2

E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}

and

E_2 = -\frac{13.6}{2^2} = -3.40\text{ eV}

Electron energy in Bohr's model depends inversely on the square of the principal quantum number.

2

Find the energy of the emitted photon

\Delta E = E_4 - E_2 = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

The energy of the emitted photon equals the energy lost during the downward transition between states.

3

Convert the photon energy from electron-volts to Joules

\Delta E = 2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

Energy must be converted to SI units (Joules) before applying Planck's equation.

4

Calculate the photon frequency using

f = \frac{\Delta E}{h}

f = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.15 \times 10^{14}\text{ Hz}

Photon energy and frequency are related by the Planck relation

E = h f

.

Anahtar Kavram

Bohr Model Energy Transitions and Photon Frequency

Atomic Models

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