The Atom and Its Properties

Atoms are stable. They combine with each other to form stable molecules or stack up to form rigid solids. They are put together systematically.

Atoms also emit and absorb light. They exist only in discrete quantum states, at which each state has a certain energy. Atoms can move from one state to another – from a higher to a lower energy level by emitting light, or from lower energy level to a higher energy level by absorbing light. The light is emitted or absorbed as a photon with energy

$$hf = E_{high} – E_{low}$$.     (1)

Atoms have angular momentum and magnetism. Every quantum state of an electron has an associated orbital angular momentum and orbital magnetic dipole moment. Every electron has a spin angular momentum and a spin magnetic dipole moment, whether it is an electron trapped in an atom or a free electron. The spin angular momentum and spin magnetic dipole moment are as intrinsic as the electron’s mass and charge.

The electron states for an atom is summarized in the table below:

Table

The angular momentum of an electron is given by the equation

$$\vec L = \vec r \times \vec p$$, in which the nucleus is the reference point.

However, the electron’s orbital angular momentum $$\vec L$$ is quantized. The allowed magnitudes of $$\vec L$$ are given by

$$L = \sqrt {\ell(\ell+1)}\hbar,\;\;\; \text{for}\;\; \ell = 0, 1, 2, … , (n-1)$$,     (2)

where $$\hbar = {h\over 2\pi}$$$$\ell$$ is the orbital quantum number, and n is the electron’s principal quantum number.

The electron can have a definite magnitude of the orbital angular momentum, but it cannot have a definite direction for it. However, definite values of a component $$L_z$$ along a chosen measurement axis can be detected or measured using the equation

$$L_z = m_{\ell}\hbar, \;\;\; \text{for}\;\;\; m_{\ell}=0, \pm 1, \pm 2, … , \pm {\ell}$$,     (3)

where $$m_{\ell}$$ is the orbital magnetic quantum number.

However, getting the definite value of $$L_z$$ will not give a definite value for $$L_x$$ and $$L_y$$. Secondly, we cannot solve separately for the definite value of each component, since getting the value of one component will change the value of the other components. Thus, we have to follow a common way of getting the allowed values of for $$L_z$$. We use the equation for the semi-classical angle $$\theta$$ given by

$$\cos \theta = {L_z \over L}$$.     (4)

The magnetic field of a magnetic dipole is set up by an orbiting charged particle. For classical particle, the dipole moment is related to the angular momentum by

$$\vec \mu_{orb} = -{e \over 2m} \vec L$$,      (5)

where $$m$$ is the mass of the particle. Since we are dealing with the atoms, the particle corresponds to the electron. The negative sign implies that the two vectors are in opposite direction, and also due to the fact that the charge of an electron is negative.

When $$\vec \mu_{orb}$$ in equation (5) is quantized, it gives the orbital magnetic dipole moment of an electron in an atom. We can get the values of the magnitude of the orbital magnetic dipole moment by substituting equation (2) to equation (5), that is

$$\mu_{orb} = {e \over 2m} \sqrt {\ell(\ell+1)\hbar}$$     (6)

in which, $$\vec \mu_{orb}$$ can have definite magnitude but cannot have definite direction. To solve for its definite value, the component on a z-axis is measured given by the equation

$$\mu_{orb,z} = -m_{\ell} {e\hbar \over 2m} = -m_{\ell} \mu_B$$,     (7)

where $$\mu_B = {eh \over 4\pi m} = {e\hbar \over 2m} = 9.274 \times 10^{-24} J/T$$ and is called as the Bohr magneton.

The electron cannot have a definite value of  $$\mu_{orb,x}$$ and $$\mu_{orb,y}$$ if it has a definite value of $$\mu_{orb,z}$$.

Every electron has a spin angular momentum \vec S or simply called as spin. It is an intrinsic angular momentum that has no classical counterpart in every electron, whether an electron is in an atom or free. The magnitude of the spin is quantized, and has values restricted to

$$S = \sqrt {s (s + 1)}\hbar,\;\;\; \text{for}\;\;\; s = \frac 12$$,       (8)

in which $$s$$ is the spin quantum number.

The spin angular momentum can have a definite magnitude but don’t have definite direction. By measuring its component on a z-axis, we could get its definite values only on that component only, and is given by

$$S_z = m_s \hbar, \;\;\; \text{for}\;\; m_s = \pm s = \pm \frac 12$$.     (9)

where m_s is the spin magnetic quantum number, which can only have two values: $$m_s = +s = + \frac 12$$, if the electron is said to be spin up, and $$m_s = -s = - \frac 12$$ if the electron is said to be spin down. If $$S_z$$ has a definite value, it follows that $$S_x$$ and $$S_y$$ cannot have definite value.

If an electron has only its intrinsic quantum numbers $$s$$ and $$m_s$$, it is a free electron. If an electron is trapped in an atom, it also has the quantum numbers $$n$$, $$\ell$$, and $$m_\ell$$.

A magnetic dipole moment is associated with the spin angular momentum. It is given by:

$$\vec \mu_s = - {e \over m} \vec S$$      (10)

The vector $$\mu_s$$ can have a definite value but it cannot have a definite direction. The definite magnitude of the vector $$\mu_s$$ is given by

$$\mu_s = {e \over m} \sqrt {s (s+1)}\hbar$$.     (11)

The vector $$\mu_s$$ can also have a definite value on the z-axis. It is given by the equation

$$\mu_{s,z} = -2m_s \mu_B$$     (12)

If $$\mu_{s,z}$$ is definite, then the vector $$\mu_s$$ cannot have a definite value $$\mu_{s,x}$$ and $$\mu_{s,y}$$.

All states with the same n from a shell. All states with the same value of n and l form a subshell. Based on the table, there are $$2(2\ell + 1)$$ states in a subshell and the total number in the shell is $$2n^2$$.

If an atom contains more than one electron, a total angular momentum $$\vec J$$ is defined. It is the vector sum of both the orbital and spin angular momenta of each electrons in the atom.

The atomic number, Z of an element is defined as the number of protons contained in the element. In an electrically neutral atom, the number of protons is equal to the number of electrons, thus, to solve for the $$\vec J$$ value of a neutral atom, we have

$$\vec J = (\vec L_1 + \vec L_2 + \vec L_3 + … + \vec L_z) + (\vec S_1 + \vec S_2 + \vec S_3 + … + \vec S_z)$$     (13)

Bohr’s Model of the Atom

Atoms are the building blocks of matter. They make up everything that we encounter every day. To describe what an atom is, several models and theories have been discovered. One of which is the one discovered by Neils Bohr.

Neils Bohr is a Danish physicist who proposed a model of the atom. He called his model the Planetary Model of an Atom. In his model, he adapted the Quantum Theory proposed in 1900 by Max Planck to the Nuclear model of an atom by Ernest Rutherford. He suggested that electrons could travel around the nucleus without radiating energy, as long as they remained in certain restricted orbits. In addition to his proposal, electron could move from one orbit to another by gaining or losing one or more quanta of energy.

He also described how the atoms emit radiation.  He proposed that an electron emits lights when it jumps from an outer orbit to an inner orbit.

Later on, his theory was used by other physicists to expand into quantum mechanics. The theory explains the structure and actions of complex atoms.