AtomsClass 12 Physics Notes

Introduction to Atomic Models

By the late 19th century, scientists knew that atoms contained negatively charged particles called electrons, which were identical across all elements. Since atoms are electrically neutral overall, they must also contain a positive charge to balance the negative charge of the electrons. The main question was: how are these positive and negative charges arranged inside an atom?

Thomson's Model of the Atom

In 1898, J. J. Thomson proposed the first model of the atom, often called the plum pudding model.

• He suggested that an atom is a sphere of uniformly distributed positive charge.
• The negatively charged electrons are embedded within this sphere, much like seeds in a watermelon or plums in a pudding.

This model, however, was later proven incorrect by further experiments.

Atomic Spectra

When we look at the light emitted by different substances, we see different types of spectra.

• Continuous Spectrum: Solids, liquids, and dense gases emit electromagnetic radiation with a continuous range of all wavelengths, like a rainbow. This is due to the complex interactions between neighbouring atoms.
• Line Spectrum (Discrete Spectrum): In contrast, rarefied (low-density) gases, when heated or electrically excited, emit light at only a few specific, discrete wavelengths. This appears as a series of bright lines on a dark background and is called an emission line spectrum. Since the atoms in these gases are far apart, the light is emitted by individual atoms.

Each element has its own unique, characteristic line spectrum, which acts like a "fingerprint." This suggested a deep connection between an atom's internal structure and the light it emits.

Rutherford's Model of the Atom

Building on Thomson's work, Ernst Rutherford conducted a famous experiment that completely changed our understanding of the atom. In 1911, his associates Hans Geiger and Ernst Marsden performed the alpha-particle scattering experiment. This led to the Rutherford's planetary model of atom, also known as the nuclear model.

According to this model:

• The entire positive charge and almost all the mass of the atom are concentrated in a very small, dense core called the nucleus.
• The electrons revolve around this nucleus in orbits, similar to how planets revolve around the sun.

While a major step forward, Rutherford's model also had problems. It couldn't explain why atoms only emit light at discrete wavelengths. According to classical physics, an orbiting electron should continuously lose energy and spiral into the nucleus, making the atom unstable.

Alpha-particle Scattering and Rutherford's Nuclear Model of Atom

The Geiger-Marsden experiment, performed at Rutherford's suggestion, was designed to probe the structure of the atom.

The Experimental Setup

1. Source: A beam of high-energy alpha-particles ($5.5 \text{ MeV}$) was obtained from a radioactive source (${ }_{83}^{214} \text{Bi}$). Alpha-particles are helium nuclei, carrying a positive charge of $+2e$.
2. Collimator: The beam was narrowed and directed by passing it through lead bricks.
3. Target: The narrow beam was aimed at a very thin foil of gold (about $2.1 \times 10^{-7} \text{ m}$ thick).
4. Detector: A rotatable detector, consisting of a zinc sulphide screen and a microscope, was used to observe the alpha-particles after they interacted with the foil. When an alpha-particle struck the screen, it produced a tiny flash of light called a scintillation.

By counting the scintillations at different angles, they could determine how the alpha-particles were being scattered by the gold atoms.

Observations and Conclusions

The results of the experiment were surprising and led to the following key observations:

• Most alpha-particles passed straight through the foil with little or no deflection. This implied that most of the atom is empty space.
• A small fraction of alpha-particles (about 0.14%) were scattered by more than $1^{\circ}$.
• A very few alpha-particles (about 1 in 8000) were deflected by more than $90^{\circ}$, with some even bouncing straight back.

Rutherford concluded that for an alpha-particle to be deflected backwards, it must have encountered a massive, concentrated positive charge. This led to the proposal of the nuclear model of the atom.

Key Features of Rutherford's Nuclear Model

• The atom consists of a tiny, dense, positively charged nucleus at its center, containing most of the atom's mass.
• The size of the nucleus is about $10^{-15} \text{ m}$ to $10^{-14} \text{ m}$, while the atom's size is about $10^{-10} \text{ m}$. This means the atom is about 10,000 to 100,000 times larger than its nucleus.
• Electrons, which are very light, orbit the nucleus at a large distance, held by the electrostatic force of attraction.
• Since most of the atom is empty space, most alpha-particles pass through undeflected. Only those that come very close to a nucleus are strongly repelled and scattered at large angles.

The force responsible for the scattering is the electrostatic (Coulomb) force between the positively charged alpha-particle ($+2e$) and the positively charged gold nucleus ($+Ze$, where Z is the atomic number). The magnitude of this repulsive force is given by: $F=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{r^{2}}$ where $r$ is the distance between the alpha-particle and the nucleus.

Alpha-particle Trajectory

The path an alpha-particle takes depends on its impact parameter (b).

• Impact Parameter (b): The perpendicular distance between the initial velocity vector of the alpha-particle and the center of the nucleus.
• Large Impact Parameter: If the particle is far from the nucleus ($b$ is large), it experiences a weak repulsive force and is deflected by a very small angle ($\theta \approx 0$).
• Small Impact Parameter: If the particle passes close to the nucleus ($b$ is small), it experiences a strong repulsive force and is scattered through a large angle.
• Head-on Collision: For a direct collision ($b \approx 0$), the particle slows down, stops, and reverses its direction, resulting in the largest possible scattering angle ($\theta \approx \pi$).

The fact that very few particles were scattered at large angles confirmed that the nucleus is extremely small.

Given

• Radius of electron's orbit, $r_{orbit} = 10^{-10} \text{ m}$
• Radius of nucleus, $r_{nucleus} = 10^{-15} \text{ m}$
• Actual radius of Earth's orbit, $R_{actual} = 1.5 \times 10^{11} \text{ m}$
• Radius of the Sun, $R_{sun} = 7 \times 10^{8} \text{ m}$

To Find

Compare the scaled-up Earth's orbit to its actual orbit.

Formula

Ratio of atomic orbit to nucleus size: $Ratio = \frac{r_{orbit}}{r_{nucleus}}$ Scaled-up Earth orbit radius: $R_{scaled} = Ratio \times R_{sun}$

Solution

First, find the ratio of the electron's orbit radius to the nucleus radius. $Ratio = \frac{10^{-10} \text{ m}}{10^{-15} \text{ m}} = 10^5$ This means the electron's orbit is $10^5$ times larger than the nucleus.

Now, apply this same ratio to the solar system. The new, scaled radius of the Earth's orbit would be $10^5$ times the radius of the Sun. $R_{scaled} = 10^5 \times (7 \times 10^8 \text{ m}) = 7 \times 10^{13} \text{ m}$

Compare this scaled radius to the actual radius of Earth's orbit: $R_{scaled} = 7 \times 10^{13} \text{ m}$ $R_{actual} = 1.5 \times 10^{11} \text{ m}$ The scaled radius is more than 100 times greater than the actual orbital radius of Earth.

Final Answer The earth would be much farther away from the sun. This implies that an atom contains a much greater fraction of empty space than our solar system does.

Given

• Kinetic energy of the $\alpha$-particle, $K = 7.7 \text{ MeV} = 7.7 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 1.2 \times 10^{-12} \text{ J}$
• Atomic number of gold, $Z = 79$
• Charge of an electron, $e = 1.6 \times 10^{-19} \text{ C}$
• Constant, $1 / 4\pi\varepsilon_0 = 9.0 \times 10^9 \text{ N m}^2/\text{C}^2$

To Find

The distance of closest approach, $d$.

Concept

At the point of closest approach, the $\alpha$-particle momentarily stops. At this instant, all of its initial kinetic energy ($K$) has been converted into electric potential energy ($U$) due to the repulsion from the gold nucleus. By the conservation of energy, $K = U$.

Formula

$K = \frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{d}$ Rearranging for $d$: $d = \frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K}$

Solution

Substitute the given values into the formula for $d$. $d = \frac{2 \times (9.0 \times 10^9 \text{ Nm}^2/\text{C}^2) \times (1.6 \times 10^{-19} \text{ C})^2 \times Z}{1.2 \times 10^{-12} \text{ J}}$ $d = \frac{(18 \times 10^9) \times (2.56 \times 10^{-38}) \times Z}{1.2 \times 10^{-12}} \text{ m}$ $d = 3.84 \times 10^{-16} Z \text{ m}$

For gold, $Z = 79$: $d(\text{Au}) = 3.84 \times 10^{-16} \times 79 \text{ m} \approx 3.0 \times 10^{-14} \text{ m}$ Since 1 fermi (fm) = $10^{-15}$ m, this is equal to 30 fm.

Final Answer The distance of closest approach is $3.0 \times 10^{-14} \text{ m}$ or 30 fm. This gives an upper limit for the radius of the gold nucleus. The $\alpha$-particle reverses its motion without ever touching the nucleus.

Electron Orbits

In Rutherford's model, for an electron to be in a stable orbit around the nucleus, the electrostatic force of attraction ($F_e$) must provide the centripetal force ($F_c$) required for circular motion.

For a hydrogen atom (one proton, one electron): $F_e = F_c$ $\frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}} = \frac{m v^{2}}{r}$

From this, we can find expressions for the energy of the electron.

• Kinetic Energy (K): $K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\varepsilon_0 r}$
• Potential Energy (U): $U = -\frac{e^2}{4\pi\varepsilon_0 r}$ (Negative because the force is attractive)
• Total Energy (E): $E = K + U = \frac{e^2}{8\pi\varepsilon_0 r} - \frac{e^2}{4\pi\varepsilon_0 r}$ $E = -\frac{e^2}{8\pi\varepsilon_0 r}$

Given

• Energy required to separate the atom (ionisation energy) = $13.6 \text{ eV}$.
• This means the total energy of the bound electron is $E = -13.6 \text{ eV}$.
• $E = -13.6 \times 1.6 \times 10^{-19} \text{ J} = -2.2 \times 10^{-18} \text{ J}$
• Mass of electron, $m = 9.1 \times 10^{-31} \text{ kg}$

To Find

• Orbital radius, $r$
• Velocity of the electron, $v$

Formula

$E = -\frac{e^2}{8\pi\varepsilon_0 r}$ $r = \frac{e^2}{4\pi\varepsilon_0 mv^2}$

Solution

First, calculate the orbital radius using the total energy formula. $r = -\frac{e^2}{8\pi\varepsilon_0 E} = -\frac{(1/4\pi\varepsilon_0)e^2}{2E}$ $r = -\frac{(9 \times 10^9 \text{ N m}^2/\text{C}^2)(1.6 \times 10^{-19} \text{ C})^2}{2(-2.2 \times 10^{-18} \text{ J})}$ $r = 5.3 \times 10^{-11} \text{ m}$

Next, calculate the velocity of the electron. We know that $K = -E$. $\frac{1}{2}mv^2 = -E$ $v = \sqrt{\frac{-2E}{m}} = \sqrt{\frac{-2(-2.2 \times 10^{-18} \text{ J})}{9.1 \times 10^{-31} \text{ kg}}}$ $v = 2.2 \times 10^6 \text{ m/s}$

Final Answer The orbital radius is $5.3 \times 10^{-11} \text{ m}$ and the velocity of the electron is $2.2 \times 10^6 \text{ m/s}$.

Bohr Model of the Hydrogen Atom

Rutherford's model was a breakthrough, but it had two major flaws based on classical physics:

1. Instability of the Atom: According to classical electromagnetic theory, an accelerating charged particle (like an electron in a circular orbit) must continuously radiate energy. This loss of energy would cause the electron to spiral inward and crash into the nucleus in a fraction of a second. This contradicts the fact that atoms are stable.
2. Continuous Spectrum: As the electron spirals inward, its frequency of revolution would change continuously. This means it should emit a continuous spectrum of light, not the discrete line spectrum that is actually observed.

Niels Bohr, in 1913, proposed a new model that addressed these problems by incorporating early quantum ideas. He introduced three revolutionary postulates.

Bohr's Postulates

1. Stable Orbits: An electron can revolve in certain specific, stable orbits (called stationary states) without emitting any radiation. Each stationary state has a definite, fixed total energy.
2. Quantisation of Angular Momentum: An electron can only exist in those orbits for which its angular momentum ($L$) is an integral multiple of $h/2\pi$, where $h$ is Planck's constant. $L = mvr = \frac{nh}{2\pi}$ Here, $n$ is an integer ($n = 1, 2, 3, ...$) called the principal quantum number.
3. Energy Transitions: An electron can "jump" from a higher energy stationary state ($E_i$) to a lower energy one ($E_f$). When it does, a single photon of light is emitted. The energy of this photon is exactly equal to the energy difference between the two states. $h\nu = E_i - E_f$ where $\nu$ is the frequency of the emitted light.

Energy Levels in the Hydrogen Atom

Using his postulates, Bohr derived formulas for the radius and energy of the allowed orbits in a hydrogen atom.

• Radius of the nth orbit: $r_n = \left(\frac{n^2}{m}\right)\left(\frac{h}{2\pi}\right)^2 \frac{4\pi\varepsilon_0}{e^2}$ This shows that the radius is proportional to $n^2$. The smallest radius ($n=1$) is called the Bohr radius.

• Energy of the nth orbit: $E_n = -\frac{me^4}{8n^2\varepsilon_0^2h^2}$ Substituting the values of the constants, this simplifies to: $E_n = -\frac{13.6}{n^2} \text{ eV}$

Understanding Energy Levels

• Ground State: The lowest possible energy state of the atom is for $n=1$. For hydrogen, this is $E_1 = -13.6 \text{ eV}$. At room temperature, most hydrogen atoms are in this state.
• Excited States: When an atom absorbs energy (e.g., from a collision or a photon), the electron can jump to a higher energy level ($n=2, 3, ...$). For example, the first excited state ($n=2$) has an energy of $E_2 = -13.6/2^2 = -3.4 \text{ eV}$.
• Ionisation Energy: The energy required to completely remove the electron from the atom (i.e., move it from $n=1$ to $n=\infty$, where $E=0$). For hydrogen, this is $0 - (-13.6 \text{ eV}) = 13.6 \text{ eV}$.

The energy level diagram (see Figure 12.7 in your textbook) visually represents these allowed energy states. The levels are not evenly spaced; they get closer together as $n$ increases.

Given

• From the previous example, for the ground state of hydrogen:
• Radius, $r = 5.3 \times 10^{-11} \text{ m}$
• Velocity, $v = 2.2 \times 10^6 \text{ m/s}$

To Find

The frequency of revolution, $\nu$.

Formula

The frequency of an object in a circular orbit is its velocity divided by the circumference of the orbit. $\nu = \frac{v}{2\pi r}$

Solution

Substitute the given values into the formula. $\nu = \frac{2.2 \times 10^6 \text{ m s}^{-1}}{2\pi(5.3 \times 10^{-11} \text{ m})}$ $\nu \approx 6.6 \times 10^{15} \text{ Hz}$

Final Answer According to classical theory, the frequency of the emitted light would be equal to the electron's frequency of revolution, which is $6.6 \times 10^{15} \text{ Hz}$. This continuous emission is what contradicts experimental observations of line spectra.

The Line Spectra of the Hydrogen Atom

Bohr's model brilliantly explained the origin of line spectra.

• Emission Lines: When an electron in an excited state ($n_i$) jumps down to a lower energy state ($n_f$), it emits a photon with a specific frequency determined by the energy difference: $h\nu = E_{n_i} - E_{n_f}$. Since only certain energy levels are allowed, only specific frequencies of light can be emitted, creating a line spectrum.
• Absorption Lines: An atom in a lower energy state can absorb a photon, but only if that photon has the exact energy needed to lift the electron to one of the allowed higher energy levels. When white light (containing all frequencies) passes through a gas, the atoms absorb photons of these specific frequencies, creating dark lines in the spectrum.

De Broglie's Explanation of Bohr's Second Postulate of Quantisation

Bohr's second postulate—that angular momentum must be quantised—was a brilliant but arbitrary rule. Why should it be so? In 1923, Louis de Broglie provided a beautiful explanation.

De Broglie proposed that particles, including electrons, have a wave-like nature. He reasoned that for an electron's orbit to be stable, its associated wave must form a standing wave.

• A standing wave is a wave that does not travel but oscillates in place. For a standing wave to form in a circular path, the circumference of the circle must be an integral multiple of the wavelength.
• Therefore, for the nth orbit: $2\pi r_n = n\lambda$ where $\lambda$ is the de Broglie wavelength of the electron.

The de Broglie wavelength is given by $\lambda = h/p$, where $p=mv$ is the momentum of the electron. Substituting this into the standing wave condition: $2\pi r_n = n \left(\frac{h}{mv_n}\right)$

Rearranging this equation gives: $mv_n r_n = \frac{nh}{2\pi}$

This is precisely Bohr's second postulate! De Broglie's hypothesis showed that the quantisation of angular momentum is a direct consequence of the wave nature of the electron. Only those orbits that can accommodate a whole number of electron wavelengths are stable.

Limitations of the Bohr Model

Despite its successes, Bohr's model was not the final picture of the atom. It had several limitations:

1. Applicable only to Hydrogenic Atoms: The model works perfectly for atoms with only one electron (like H, He$^{+}$, Li$^{2+}$) but fails for atoms with more than one electron because it doesn't account for the repulsive forces between electrons.
2. Cannot Explain Spectral Line Intensities: The model correctly predicts the frequencies of spectral lines but cannot explain why some lines are brighter (more intense) than others.
3. Inconsistent with Quantum Mechanics: It combines classical ideas (defined orbits) with quantum postulates, which is not fully consistent. Later developments in quantum mechanics replaced the idea of fixed orbits with probability clouds (orbitals).