Subatomic Particles and Quantum Theory
Subatomic Particles of the Atom
In 1886, German physicist E. Goldstein discovered canal rays using a discharge tube fitted with a perforated cathode. When high-speed cathode rays (electrons) strike gas molecules inside the tube, they knock out electrons from those molecules, producing positive ions. These positive ions move toward the cathode and some pass through the fine holes, creating a glow on the wall opposite the anode. Since these rays travel through canals in the cathode, they are called canal rays, and because they carry a positive charge, they are also called positive rays. The reaction producing these ions is:
$$M + e^- \rightarrow M^+ + 2e^-$$
Goldstein's Experiment: A discharge tube with a cathode containing fine perforations was used. When high voltage was applied, both cathode rays (moving away from cathode) and positive rays (moving toward cathode through the perforations) were produced simultaneously.
Origin of Positive Rays: The positive ions are created when fast-moving electrons from the cathode collide with and ionize neutral gas molecules in the tube.
Identification of the Proton: When hydrogen gas is used, the positive particle obtained has the highest charge-to-mass ratio of any gas, confirming it as the lightest positive particle. Rutherford named this particle the proton.
Mass Comparison: The mass of a proton is 1836 times the mass of an electron, making it the lightest among all positively charged particles obtained from different gases.
Properties of Positive Rays
•
Deflected by electric and magnetic fields (positive charge)
•
Travel in straight lines opposite to cathode rays
•
Produce flashes on a ZnS plate
•
e/m value depends on the gas used — heavier gas gives smaller e/m
•
The e/m value is always smaller than that of electrons
An atom is composed of three fundamental subatomic particles: the proton, the neutron, and the electron. Protons and neutrons are concentrated in the tiny central nucleus, while electrons occupy regions of space around it. Each particle has distinct charge and mass properties that define atomic structure and chemical behaviour.
Proton: Carries a relative charge of +1 and a mass of approximately 1.0073 amu. It is found inside the nucleus and determines the atomic number of an element.
Neutron: Carries no charge (electrically neutral) and has a mass of approximately 1.0087 amu, slightly heavier than a proton. It also resides in the nucleus and contributes to atomic mass.
Electron: Carries a relative charge of -1 and has a mass of approximately $5.486 \times 10^{-4}$ amu — about 1836 times lighter than a proton. Electrons occupy the space outside the nucleus.
Fundamental Particle Properties
•
Proton: charge = +1.6022 × 10⁻¹⁹ C, mass = 1.6726 × 10⁻²⁷ kg (1.0073 amu)
•
Neutron: charge = 0, mass = 1.6750 × 10⁻²⁷ kg (1.0087 amu)
•
Electron: charge = -1.6022 × 10⁻¹⁹ C, mass = 9.1095 × 10⁻³¹ kg (5.486 × 10⁻⁴ amu)
Key Mass Relationships
•
Mass of proton ≈ 1836 × mass of electron
•
Mass of neutron ≈ mass of proton (slightly greater)
•
Mass of electron ≈ 1/1836 of proton mass
Planck's Quantum Theory and the Photon
In 1900, Max Planck proposed the quantum theory to explain the emission and absorption of radiation. According to this revolutionary theory, energy is not emitted or absorbed continuously. Instead, it travels in discrete packets called quantuma (plural: quanta). In the case of light, each quantum of energy is called a photon. The energy associated with each photon is directly proportional to its frequency.
$$E = h\nu$$
Relates the energy of a photon to its frequency — higher frequency radiation carries more energy per photon
$E$=Energy of one photon(joules (J))
$h$=Planck's constant(6.626 × 10⁻³⁴ J·s)
$\nu$=Frequency of the radiation(hertz (Hz, s⁻¹))
Energy is Quantized: A body can emit or absorb energy only in whole-number multiples of the quantum. Fractional quanta cannot exist.
Photon Concept: The photon is the smallest discrete unit of electromagnetic energy. It is massless and always travels at the speed of light.
Planck's Constant: The constant $h = 6.626 \times 10^{-34}$ J·s is the ratio of energy to frequency for any photon. It is a fundamental constant of nature.
Frequency Dependence: The energy of a photon increases with increasing frequency. Blue light (higher frequency) carries more energy per photon than red light (lower frequency).
The energy of a photon can be expressed in three equivalent forms depending on which property of the radiation is known. The frequency $\nu$, wavelength $\lambda$, and wave number $\bar{\nu}$ are all interrelated through the speed of light. The wavelength is the distance between two adjacent crests (or troughs) of a wave, and the wave number is the number of waves per unit length — the reciprocal of wavelength.
$$E = \frac{hc}{\lambda}$$
Relates photon energy to wavelength — longer wavelength means lower energy per photon
$E$=Energy of one photon(joules (J))
$h$=Planck's constant(6.626 × 10⁻³⁴ J·s)
$c$=Speed of light(3.0 × 10⁸ m/s)
$\lambda$=Wavelength of the radiation(metres (m))
$\lambda$ is large (radio waves)
→Photon energy is very small
$\lambda$ is small (gamma rays)
→Photon energy is very large
Frequency–Wavelength Link: The frequency and wavelength of any electromagnetic wave are related by $\nu = c / \lambda$. Doubling the frequency halves the wavelength.
Wave Number Form: The wave number $\bar{\nu} = 1/\lambda$ gives energy as $E = hc\bar{\nu}$. Greater wave number means greater photon energy.
Inverse Relationships: Greater wavelength → smaller frequency → lower energy. Greater wave number → smaller wavelength → higher energy.
Wavelength Units: Common units are Å ($10^{-10}$ m), nm ($10^{-9}$ m), and pm ($10^{-12}$ m). Always convert to metres for calculations with SI constants.
Energy Relationships Summary
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$E = h\nu$ (from frequency)
•
$E = hc/\lambda$ (from wavelength)
•
$E = hc\bar{\nu}$ (from wave number)
•
$\bar{\nu} = 1/\lambda$ (wave number from wavelength)
Quantum Numbers and Energy Levels
An electron in an atom is completely described by four quantum numbers. Three of these arise from solving the Schrödinger wave equation for the hydrogen atom, and the fourth (spin) was added experimentally. These four numbers define the shell, subshell, orbital orientation, and spin direction of an electron.
Principal Quantum Number ($n$): Defines the shell or main energy level. Takes integer values $n = 1, 2, 3, ...$. The first shell ($n=1$) is nearest to the nucleus and has the lowest energy. Maximum electrons in a shell = $2n^2$.
Azimuthal Quantum Number ($l$): Defines the subshell and the shape of the orbital. Takes values $l = 0, 1, 2, ..., n-1$. Each value corresponds to a subshell type: $l = 0$ (s), $l = 1$ (p), $l = 2$ (d), $l = 3$ (f).
Magnetic Quantum Number ($m_l$): Defines the orientation of the orbital in space. Takes integer values from $-l$ to $+l$, including zero. For a p-subshell ($l=1$), $m_l = -1, 0, +1$ (three orbitals).
Spin Quantum Number ($m_s$): Describes the direction of electron spin. Can only be $+1/2$ (clockwise, ↑) or $-1/2$ (anticlockwise, ↓). Two electrons in the same orbital must have opposite spins.
Quantum Number Values for Common Subshells
1
1s: $n=1,\; l=0,\; m_l=0$ — 1 orbital
2
2s: $n=2,\; l=0,\; m_l=0$ — 1 orbital
3
2p: $n=2,\; l=1,\; m_l=-1,\; 0,\; +1$ — 3 orbitals
4
3d: $n=3,\; l=2,\; m_l=-2,\; -1,\; 0,\; +1,\; +2$ — 5 orbitals
5
4f: $n=4,\; l=3,\; m_l=-3,\; -2,\; -1,\; 0,\; +1,\; +2,\; +3$ — 7 orbitals
The energy of a subshell depends on both the principal quantum number $n$ and the azimuthal quantum number $l$. The $(n+l)$ rule states that subshells are filled in order of increasing $(n+l)$ value. If two subshells have the same $(n+l)$ value, the one with the smaller $n$ is filled first. This gives the energy ordering: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s.
$$\text{Maximum electrons in shell} = 2n^2$$
Gives the total number of electrons that can be accommodated in a shell with principal quantum number $n$
$n$=Principal quantum number (shell number)(dimensionless integer)
$2n^2$=Total electron capacity of the shell(electrons)
$n = 1$
→Maximum 2 electrons (K shell)
$n = 2$
→Maximum 8 electrons (L shell)
$n = 3$
→Maximum 18 electrons (M shell)
$n = 4$
→Maximum 32 electrons (N shell)
(n+l) Rule: Calculate $n + l$ for each subshell. Lower $(n+l)$ means lower energy and is filled first. For equal $(n+l)$, lower $n$ is preferred.
Energy Ordering: 4s ($n+l=4$) fills before 3d ($n+l=5$) despite 3d having a lower principal quantum number. This is why transition metals fill the 4s orbital before 3d.
Subshell Capacities: s-subshells hold 2 electrons (1 orbital), p-subshells hold 6 (3 orbitals), d-subshells hold 10 (5 orbitals), f-subshells hold 14 (7 orbitals).
Shell Names: $n=1$ (K), $n=2$ (L), $n=3$ (M), $n=4$ (N). The K shell is closest to the nucleus and has the lowest energy.
Subshell Energy Ordering by (n+l) Rule
1
1s ($1+0=1$)
2
2s ($2+0=2$)
3
2p ($2+1=3$)
4
3s ($3+0=3$)
5
3p ($3+1=4$), 4s ($4+0=4$) — same sum, 3s before 4s
6
3d ($3+2=5$), 4p ($4+1=5$)
7
4d ($4+2=6$), 5s ($5+0=5$), 5p ($5+1=6$)
Three rules govern how electrons are distributed among orbitals within subshells. Together with the $(n+l)$ energy ordering, these rules allow the complete electronic configuration of any element to be written.
Aufbau Principle: Electrons fill subshells in order of increasing energy. The lowest-energy subshell available is always filled before moving to the next higher one.
Pauli Exclusion Principle: No two electrons in the same orbital can have all four quantum numbers identical. Consequently, each orbital holds a maximum of two electrons, and these must have opposite spins (↑↓).
Hund's Rule: When electrons are placed in degenerate orbitals (orbitals of the same energy within a subshell), they occupy separate orbitals with parallel spins before any orbital is doubly occupied. For carbon, the 2p electrons enter two different p-orbitals with the same spin rather than pairing in one orbital.
Orbitals and Their Shapes
An orbital is a three-dimensional region around the nucleus where there is a high probability of finding an electron. The shape of an orbital is determined by the azimuthal quantum number $l$. The probability of finding the electron at a particular point is given by $|\psi|^2$, where $\psi$ is the wave function from Schrödinger's equation. Unlike the fixed circular paths (orbits) in Bohr's model, orbitals represent probability distributions.
s-Orbitals ($l = 0$): Spherically symmetrical around the nucleus. The probability of finding an electron depends only on the distance from the nucleus, not the direction. All s-orbitals have this same spherical shape, but higher $n$ values give larger spheres.
p-Orbitals ($l = 1$): Dumbbell-shaped with a node (zero electron density) at the nucleus. There are three p-orbitals oriented along the three coordinate axes: $p_x$ (along x-axis), $p_y$ (along y-axis), and $p_z$ (along z-axis). These three are degenerate orbitals — same energy, different orientation.
d-Orbitals ($l = 2$): Five orbitals with more complex shapes: $d_{xy}$ (lobes between x and y axes), $d_{yz}$ (between y and z axes), $d_{xz}$ (between x and z axes), $d_{x^2-y^2}$ (lobes along x and y axes), and $d_{z^2}$ (two lobes along z-axis with a doughnut-shaped region in the xy-plane).
f-Orbitals ($l = 3$): Seven orbitals with even more complex shapes. These are relevant for lanthanides and actinides but are less commonly needed at this level.
Orbital Shapes at a Glance
1
s-orbital: spherical, 1 per subshell, 2 electrons max
2
p-orbital: dumbbell, 3 per subshell ($p_x, p_y, p_z$), 6 electrons max
3
d-orbital: five distinct shapes, 5 per subshell, 10 electrons max
4
f-orbital: seven complex shapes, 7 per subshell, 14 electrons max
Every orbital contains regions where the probability of finding an electron is zero. These are called nodes. The number and type of nodes depend on the quantum numbers $n$ and $l$. Understanding nodes is essential for visualising how orbitals change with increasing energy. The total number of nodes increases with the principal quantum number, meaning higher-energy orbitals have more complex shapes.
$$\text{Number of orbitals in a subshell} = 2l + 1$$
Gives the number of distinct orbitals within a subshell based on its azimuthal quantum number
$l$=Azimuthal quantum number(dimensionless integer)
$2l + 1$=Number of orbitals in the subshell(orbitals)
$l = 0$ (s-subshell)
→1 orbital
$l = 1$ (p-subshell)
→3 orbitals
$l = 2$ (d-subshell)
→5 orbitals
$l = 3$ (f-subshell)
→7 orbitals
Radial Nodes: Spherical surfaces where electron probability drops to zero. The number of radial nodes equals $n - l - 1$. For example, the 2s orbital ($n=2, l=0$) has $2 - 0 - 1 = 1$ radial node.
Angular Nodes: Planes or cones where electron probability is zero. The number of angular nodes equals $l$. All s-orbitals have zero angular nodes (hence their spherical symmetry), p-orbitals have one, and d-orbitals have two.
Total Nodes: The total number of nodes in any orbital is $n - 1$. This is the sum of radial and angular nodes: $(n - l - 1) + l = n - 1$.
Electron Capacity: Each orbital holds at most 2 electrons (Pauli exclusion). The total electrons in a subshell is $2(2l + 1)$. The total electrons in a shell is $2n^2$.
Node Counting for Selected Orbitals
1
1s ($n=1, l=0$): 0 radial + 0 angular = 0 total nodes
2
2s ($n=2, l=0$): 1 radial + 0 angular = 1 total node
3
2p ($n=2, l=1$): 0 radial + 1 angular = 1 total node
4
3s ($n=3, l=0$): 2 radial + 0 angular = 2 total nodes
5
3p ($n=3, l=1$): 1 radial + 1 angular = 2 total nodes
6
3d ($n=3, l=2$): 0 radial + 2 angular = 2 total nodes
Application Strategy
Worked Examples
Problem Statement
Calculate the energy of a photon with a frequency of $6.0 \times 10^{14}$ Hz. ($h = 6.626 \times 10^{-34}$ J·s)
Planck's Quantum Theoryphoton
Strategic Toolkit
$$E = h\nu$$
Step 1
Identify the formula
Use Planck's equation which relates photon energy directly to frequency.
$$E = h\nu$$
Step 2
Substitute the known values
Insert Planck's constant and the given frequency into the equation.
$$E = (6.626 \times 10^{-34}) \times (6.0 \times 10^{14})$$
Step 3
Calculate the energy
Multiply the values to obtain the energy of a single photon in joules.
$$E = 3.98 \times 10^{-19} \text{ J}$$
Problem Statement
A photon has an energy of $4.97 \times 10^{-19}$ J. Calculate its wavelength in nanometres. ($h = 6.626 \times 10^{-34}$ J·s, $c = 3.0 \times 10^8$ m/s)
photonwavelength
Strategic Toolkit
$$E = \frac{hc}{\lambda}$$
Step 1
Rearrange for wavelength
Solve the energy-wavelength equation for $\lambda$ by rearranging algebraically.
$$\lambda = \frac{hc}{E}$$
Step 2
Substitute values
Insert the values of $h$, $c$, and the given photon energy into the rearranged formula.
$$\lambda = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{4.97 \times 10^{-19}}$$
Step 3
Calculate and convert units
Evaluate the expression to get the wavelength in metres, then convert to nanometres by multiplying by $10^9$.
$$\lambda = 4.0 \times 10^{-7} \text{ m} = 400 \text{ nm}$$
Problem Statement
Light has a wavelength of 500 nm. Calculate its wave number in m⁻¹.
wave numberwavelength
Strategic Toolkit
$$\bar{\nu} = \frac{1}{\lambda}$$
Step 1
Convert wavelength to metres
Convert nanometres to metres since wave number in m⁻¹ requires the wavelength in metres.
$$\lambda = 500 \times 10^{-9} \text{ m} = 5.0 \times 10^{-7} \text{ m}$$
Step 2
Apply the wave number formula
Take the reciprocal of the wavelength to find the wave number.
$$\bar{\nu} = \frac{1}{5.0 \times 10^{-7}}$$
Step 3
Calculate the result
Compute the final wave number value.
$$\bar{\nu} = 2.0 \times 10^{6} \text{ m}^{-1}$$
Problem Statement
Write all four quantum numbers for each of the first two electrons entering the 3d subshell.
quantum numbersubshell
Strategic Toolkit
$$m_l = -l, ..., +l$$
Step 1
Identify n and l from the subshell label
The number 3 in 3d gives $n = 3$, and the letter d corresponds to $l = 2$.
$$n = 3, \quad l = 2$$
Step 2
Determine possible $m_l$ values
For $l = 2$, the magnetic quantum number ranges from $-2$ to $+2$, giving five possible orbitals.
$$m_l = -2, \; -1, \; 0, \; +1, \; +2$$
Step 3
Apply Hund's rule for the first two electrons
The first two electrons enter separate degenerate orbitals with parallel spins. By convention, we fill starting from the lowest $m_l$ value.
$$\text{Electron 1: } (3, 2, -2, +\tfrac{1}{2}) \qquad \text{Electron 2: } (3, 2, -1, +\tfrac{1}{2})$$
Problem Statement
Arrange the following subshells in order of increasing energy: 4s, 3d, 4p, 3p.
subshellprincipal quantum number
Strategic Toolkit
$$(n+l) \text{ rule}$$
Step 1
Calculate (n+l) for each subshell
Compute the sum of principal and azimuthal quantum numbers for each subshell.
$$\text{3p: } 3+1=4, \quad \text{4s: } 4+0=4, \quad \text{3d: } 3+2=5, \quad \text{4p: } 4+1=5$$
Step 2
Apply the tiebreaker for equal (n+l)
When two subshells have the same $n+l$ value, the one with the smaller $n$ is filled first. Both 3p and 4s have $n+l=4$, so 3p (smaller $n$) comes before 4s. Similarly, 3d comes before 4p.
$$3p \; (n=3) < 4s \; (n=4), \qquad 3d \; (n=3) < 4p \; (n=4)$$
Step 3
Write the final energy order
Combine the groups to give the complete arrangement from lowest to highest energy.
$$\text{3p} < \text{4s} < \text{3d} < \text{4p}$$
Memorize formulas
Photon Energy from Frequency
Use when frequency is given. $h = 6.626 \times 10^{-34}$ J·s, $\nu$ in Hz, $E$ in joules.
Formula
$$E = h\nu$$
Photon Energy from Wavelength
Use when wavelength is given. $c = 3.0 \times 10^8$ m/s, $\lambda$ must be in metres, $E$ in joules.
Formula
$$E = \frac{hc}{\lambda}$$
Photon Energy from Wave Number
Use when wave number is given. $\bar{\nu}$ in m⁻¹, $E$ in joules.
Formula
$$E = hc\bar{\nu}$$
Maximum Electrons in a Shell
Gives the total electron capacity of shell $n$. For K shell ($n=1$): 2; L shell ($n=2$): 8; M shell ($n=3$): 18.
Formula
$$\text{Max electrons} = 2n^2$$
Number of Orbitals in a Subshell
Use the azimuthal quantum number $l$ of the subshell. s ($l=0$): 1; p ($l=1$): 3; d ($l=2$): 5; f ($l=3$): 7.
Formula
$$\text{Number of orbitals} = 2l + 1$$
Review
1
Cathode rays are deflected towards the positive plate in an electric field. This confirms that cathode rays consist of:
Ans
Negatively charged particles
Review
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Identify the particle
Cathode rays were identified by J.J. Thomson as streams of electrons — fundamental particles with a negative charge.
Electron: charge = $-1.602 \times 10^{-19}$ C
Apply electric field behaviour
A negatively charged particle in an electric field is attracted to the positive plate and repelled by the negative plate.
Deflection direction confirms negative charge
2
Canal rays, which led to the discovery of the proton, were first observed by:
Ans
Eugen Goldstein
Review
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Canal ray discovery
Goldstein used a discharge tube with a perforated cathode. Positive ions passed through the holes and produced glowing rays behind the cathode.
Canal rays = streams of positive ions (protons)
Historical context
Goldstein's work predated the formal identification of the proton. The proton was later characterised by Rutherford in 1917.
Goldstein (1886) → Rutherford (1917 proton identification)
3
Which of the following statements correctly describes a proton?
Ans
It carries a positive charge of $+1.602 \times 10^{-19}$ C
Review
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Proton charge
The proton has a fundamental positive charge equal to the elementary charge.
$q_p = +1.602 \times 10^{-19}$ C
Proton mass
The proton mass is $1.673 \times 10^{-27}$ kg, roughly 1836 times the electron mass of $9.109 \times 10^{-31}$ kg.
$m_p \approx 1836 \times m_e$
4
The neutron was discovered in 1932. What is its approximate mass?
Ans
$1.675 \times 10^{-27}$ kg
Review
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Neutron mass
James Chadwick discovered the neutron in 1932 by bombarding beryllium with alpha particles.
$m_n = 1.675 \times 10^{-27}$ kg
Comparison
The neutron mass is nearly identical to the proton mass, differing by only about 0.14%.
$m_n \approx m_p$ (both nuclear constituents)
5
A photon is best described as:
Ans
A quantum of electromagnetic radiation
Review
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Photon definition
According to Planck's quantum theory, electromagnetic energy is emitted or absorbed in discrete packets called photons (quanta).
Photon = smallest unit of electromagnetic energy
Energy of a photon
The energy of a photon is directly proportional to its frequency.
$E = h\nu$
6
The energy of a photon with a frequency of $5.0 \times 10^{14}$ Hz is:
Ans
$3.31 \times 10^{-19}$ J
Review
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Identify the formula
When frequency is given, use the direct photon energy formula.
$E = h\nu$
Substitute values
$E = (6.626 \times 10^{-34} \text{ J·s})(5.0 \times 10^{14} \text{ Hz})$
$E = 3.313 \times 10^{-19}$ J
Convert to electron-volts (optional)
Divide by the elementary charge to get energy in eV.
$E = \frac{3.313 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 2.07$ eV
7
A photon has an energy of $5.52 \times 10^{-19}$ J. Its wavelength is closest to:
Ans
360 nm
Review
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Rearrange the energy-wavelength formula
Starting from $E = \frac{hc}{\lambda}$, solve for $\lambda$.
$\lambda = \frac{hc}{E}$
Substitute values
$\lambda = \frac{(6.626 \times 10^{-34})(3.0 \times 10^8)}{5.52 \times 10^{-19}}$
$\lambda = 3.60 \times 10^{-7}$ m $= 360$ nm
8
The wave number of light with a wavelength of 500 nm is:
Ans
$2.0 \times 10^{6}$ m$^{-1}$
Review
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Convert wavelength to SI units
Convert nanometres to metres for consistent SI calculation.
$\lambda = 500 \times 10^{-9}$ m $= 5.0 \times 10^{-7}$ m
Apply wave number definition
Wave number is the reciprocal of wavelength.
$\bar{\nu} = \frac{1}{5.0 \times 10^{-7}} = 2.0 \times 10^{6}$ m$^{-1}$
9
The principal quantum number (n) describes:
Ans
The energy level (shell) in which an electron is found
Review
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Principal quantum number
n can take integer values 1, 2, 3, … and determines the shell (K, L, M, …).
$n = 1, 2, 3, \ldots$
Properties governed by n
Higher n means the electron occupies a shell further from the nucleus with higher energy.
Shell size and energy increase with n
10
For the principal quantum number $n = 3$, how many subshells are present and what are their azimuthal quantum number values?
Ans
3 subshells: l = 0, 1, 2
Review
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Determine the range of l
For any given n, the azimuthal quantum number l takes values from 0 to $n - 1$.
$l = 0, 1, 2, \ldots, n - 1$
Apply to n = 3
$l$ can be 0, 1, or 2, giving three subshells: 3s, 3p, 3d.
3 subshells: $l = 0, 1, 2$
11
How many orbitals are present in the 4f subshell?
Ans
7
Review
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Identify the azimuthal quantum number
For an f subshell, $l = 3$.
$l = 3$ (4f subshell)
Apply the orbital count formula
Number of orbitals $= 2l + 1 = 2(3) + 1$.
$7$ orbitals in the 4f subshell
12
The maximum number of electrons that can be accommodated in the shell with $n = 4$ is:
Ans
32
Review
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Apply the shell capacity formula
The maximum number of electrons in a shell is given by $2n^2$.
$\text{Max} = 2n^2$
Substitute n = 4
$\text{Max} = 2(4)^2 = 2 \times 16 = 32$
32 electrons in the $n = 4$ shell
13
Which statement correctly describes an s orbital?
Ans
It is spherical in shape with no angular nodes
Review
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Determine orbital shape from l
For $l = 0$ (s orbital), the angular node count is $l = 0$, meaning no nodal planes.
Angular nodes $= l = 0$ → spherical shape
Electron capacity of a single orbital
Any orbital can hold at most 2 electrons with opposite spins ($m_s = +\frac{1}{2}$ and $m_s = -\frac{1}{2}$).
Maximum 2 electrons per orbital
14
The number of angular nodes in any orbital is equal to:
Ans
The azimuthal quantum number $l$
Review
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Angular nodes formula
Angular nodes are regions where the probability of finding an electron is zero due to angular dependence.
Angular nodes $= l$
Total nodes relationship
Total nodes $= n - 1$. Radial nodes $= n - 1 - l$.
Radial nodes $= (n - 1) - l$
15
According to the $(n + l)$ rule, which of the following subshells is filled first?
Ans
4s
Review
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Calculate n + l for each subshell
4s: $4 + 0 = 4$; 3d: $3 + 2 = 5$; 4p: $4 + 1 = 5$; 5s: $5 + 0 = 5$
4s ($n + l = 4$) < 3d, 4p, 5s ($n + l = 5$)
Apply the filling order rule
Subshells with lower $n + l$ are filled first. This explains why 4s fills before 3d despite 3d having a lower principal quantum number.
Filling order: 4s → 3d → 4p → 5s
Question 1 of 15
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1
Cathode rays are deflected towards the positive plate in an electric field. This confirms that cathode rays consist of: