Step 1: Establishing the Energy Expression for Hydrogen Atom

The energy of an electron in the $n^{th}$ Bohr orbit of a hydrogen atom is given by the Rydberg-Bohr formula:

$E_n = -\frac{R_H}{n^2}$

Given the energy of the ground state ($n=1$) is $-2.18 \times 10^{-11} \text{ ergs}$, we first convert this value to Joules ($1 \text{ erg} = 10^{-7} \text{ J}$):

$E_1 = -2.18 \times 10^{-11} \times 10^{-7} \text{ J} = -2.18 \times 10^{-18} \text{ J}$ [Per definition of SI units conversion].

Step 2: Calculating Energy Change for Transition ($n_1 = 1$ to $n_2 = 5$)

The energy required ($\Delta E$) to shift the electron from $n_1$ to $n_2$ is calculated by the difference in energy levels [Per Bohr's Postulate]:

$\Delta E = E_5 - E_1 = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Substituting $R_H = 2.18 \times 10^{-18} \text{ J}$ and the quantum numbers:

$\Delta E = 2.18 \times 10^{-18} \left( \frac{1}{1^2} - \frac{1}{5^2} \right) \text{ J}$

$\Delta E = 2.18 \times 10^{-18} \left( 1 - 0.04 \right) \text{ J}$

$\Delta E = 2.18 \times 10^{-18} \times 0.96 \text{ J} = 2.0928 \times 10^{-18} \text{ J}$

Step 3: Determining the Wavelength of Emitted Light

When the electron returns from $n=5$ to $n=1$, the energy released is equal to the energy absorbed: $\Delta E = 2.0928 \times 10^{-18} \text{ J}$. The relationship between energy and wavelength ($\lambda$) is given by the Planck-Einstein relation [Per Planck’s Quantum Theory]:

$E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$

Constants:

• $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• $c = 3.0 \times 10^8 \text{ m/s}$

$\lambda = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3.0 \times 10^8 \text{ m/s})}{2.0928 \times 10^{-18} \text{ J}}$

$\lambda = \frac{19.878 \times 10^{-26}}{2.0928 \times 10^{-18}} \text{ m}$

$\lambda = 9.50 \times 10^{-8} \text{ m}$

Visual Representation of Transition

Final Solution: The energy required is $2.0928 \times 10^{-18} \text{ J}$, and the wavelength of the emitted light is $9.50 \times 10^{-8} \text{ m}$ (or $95.0 \text{ nm}$).

Step 1: Identified Variables and Constants

To determine the de Broglie wavelength ($\lambda$), we utilize the de Broglie relation. The following physical constants and given variables are defined:

• Velocity of the electron ($v$): $1.6 \times 10^6 \, \text{m s}^{-1}$
• Mass of an electron ($m_e$): $9.109 \times 10^{-31} \, \text{kg}$ [Accepted physical constant]
• Planck’s constant ($h$): $6.626 \times 10^{-34} \, \text{J s}$ (or $\text{kg m}^2 \text{s}^{-1}$)

Step 2: Theoretical Framework

Per the de Broglie hypothesis, the wavelength ($\lambda$) of a moving particle is inversely proportional to its momentum ($p$). The relationship is defined as:

$$\lambda = \frac{h}{mv} = \frac{h}{p}$$

[Justification: This is derived from the synthesis of Einstein's mass-energy equivalence $E=mc^2$ and Planck's quantum theory $E=h\nu$, establishing the wave-particle duality of matter.]

Step 3: Execution of the Calculation

Substitute the constants and variables into the de Broglie equation:

$$\lambda = \frac{6.626 \times 10^{-34} \, \text{kg m}^2 \text{s}^{-1}}{(9.109 \times 10^{-31} \, \text{kg}) \times (1.6 \times 10^6 \, \text{m s}^{-1})}$$

Calculating the denominator (momentum, $p$):

$$p = 9.109 \times 1.6 \times 10^{-31 + 6} = 14.5744 \times 10^{-25} \, \text{kg m s}^{-1}$$

Calculating the final wavelength ($\lambda$):

$$\lambda = \frac{6.626 \times 10^{-34}}{14.5744 \times 10^{-25}}$$

$$\lambda \approx 0.4546 \times 10^{-9} \, \text{m}$$

Step 4: Dimensional and Unit Verification

Expression	Resulting Units
$\frac{\text{kg m}^2 \text{s}^{-1}}{\text{kg} \times \text{m s}^{-1}}$	$\text{m}$ (Meters)

Final Solution: The de Broglie wavelength associated with the electron is $4.546 \times 10^{-10} \, \text{m}$ (or $0.4546 \, \text{nm}$).

Step 1: Determining Total Electrons in Neutral Atoms

To determine the total number of electrons in molecular species, we first identify the atomic number ($Z$) of the constituent atoms, which represents the number of electrons in a neutral atom [Per the definition of the Periodic Table and Atomic Structure].

• Hydrogen ($H$): $Z = 1$
• Oxygen ($O$): $Z = 8$

Step 2: Calculating Electrons for $H_2^+$

The species $H_2^+$ consists of two hydrogen atoms with a net positive charge of $+1$.

• Total electrons in two neutral $H$ atoms: $2 \times 1 = 2$
• Effect of positive charge ($+$): Subtract 1 electron from the total.
• Calculation: $2 - 1 = 1$ electron.

Step 3: Calculating Electrons for $H_2$

The species $H_2$ is a neutral diatomic hydrogen molecule.

• Total electrons in two neutral $H$ atoms: $2 \times 1 = 2$
• Calculation: $2$ electrons.

Step 4: Calculating Electrons for $O_2^+$

The species $O_2^+$ consists of two oxygen atoms with a net positive charge of $+1$.

• Total electrons in two neutral $O$ atoms: $2 \times 8 = 16$
• Effect of positive charge ($+$): Subtract 1 electron from the total.
• Calculation: $16 - 1 = 15$ electrons.

Summary Table of Electronic Counts

Species	Calculation	Total Electrons
$H_2^+$	$(2 \times 1) - 1$	1
$H_2$	$(2 \times 1)$	2
$O_2^+$	$(2 \times 8) - 1$	15

Visual Representation of Molecular Electron Distribution

Final Solution: The number of electrons are: $H_2^+ = 1$, $H_2 = 2$, and $O_2^+ = 15$.

Step 1: Analyzing the Noble Gas Core

The notation $[Ne]$ represents the electron configuration of Neon, which is the noble gas preceding the element in question. The atomic number of Neon ($Z$) is $10$.

The full configuration for Neon is $1s^2 2s^2 2p^6$. [Per the Aufbau Principle, electrons fill lower energy orbitals before higher energy ones].

Step 2: Calculating Total Electron Count

The given configuration is $[Ne] 3s^2 3p^3$. To find the total number of electrons ($N_e$), we sum the electrons in the Neon core and the valence shell:

$N_e = (\text{Electrons in } Ne) + (\text{Electrons in } 3s) + (\text{Electrons in } 3p)$

$N_e = 10 + 2 + 3 = 15$

[By the Principle of Electroneutrality, in a neutral atom, the number of electrons equals the atomic number $Z$].

Step 3: Determining the Element

Since the total number of electrons is $15$, the atomic number of the element is $Z = 15$.

Consulting the Periodic Table of Elements for $Z = 15$, we identify the element as Phosphorus ($P$).

Visual Representation of Valence Configuration

Summary of Electronic Distribution

Orbital	Number of Electrons
$1s$	$2$
$2s$	$2$
$2p$	$6$
$3s$	$2$
$3p$	$3$
Total	$15$

Final Solution: The atom indicated by the configuration $[Ne] 3s^2 3p^3$ is Phosphorus (P), with an atomic number of 15.

Step 1: Given Variables & Physical Constants

To determine the energy of a photon with a specific wavelength, we utilize the fundamental relationship between energy, frequency, and wavelength established by Planck’s Quantum Theory [Per Planck's Law].

• Wavelength ($\lambda$): $0.50 \text{ Å} = 0.50 \times 10^{-10} \text{ m}$ [Conversion: $1 \text{ Å} = 10^{-10} \text{ m}$]
• Planck’s Constant ($h$): $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Speed of Light ($c$): $3.0 \times 10^8 \text{ m/s}$

Step 2: Mathematical Formulation

The energy ($E$) of a single photon is defined by the equation:

$E = \frac{hc}{\lambda}$

[Justification: Since frequency ($\nu$) is given by $\nu = \frac{c}{\lambda}$, substituting this into $E = h\nu$ yields the formula above.]

Step 3: Sequential Calculation

Substitute the constants into the derived equation:

$E = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3.0 \times 10^8 \text{ m/s})}{0.50 \times 10^{-10} \text{ m}}$

Multiply the numerator:

$E = \frac{19.878 \times 10^{-26} \text{ J}\cdot\text{m}}{0.50 \times 10^{-10} \text{ m}}$

Perform the division:

$E = 39.756 \times 10^{-16} \text{ J}$

Adjusting to standard scientific notation:

$E = 3.9756 \times 10^{-15} \text{ J}$

Visual Representation of Photon Energy vs Wavelength

Step 4: Summary of Units and Significant Figures

The given wavelength $0.50 \text{ Å}$ consists of two significant figures. Therefore, the result should be expressed accordingly.

Variable	Value
Energy (Exact)	$3.9756 \times 10^{-15} \text{ J}$
Energy (Rounded)	$4.0 \times 10^{-15} \text{ J}$

Final Solution: The energy of each photon is $3.98 \times 10^{-15} \text{ J}$ (or $4.0 \times 10^{-15} \text{ J}$ to two significant figures).

Step 1: Given Variables and Fundamental Constants

• Work function ($\Phi$) = $1.9 \text{ eV}$
• Wavelength of incident radiation ($\lambda$) = $500 \text{ nm} = 500 \times 10^{-9} \text{ m}$
• Planck’s constant ($h$) = $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Speed of light ($c$) = $3 \times 10^8 \text{ m/s}$
• Conversion factor: $1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}$
• Mass of electron ($m_e$) = $9.11 \times 10^{-31} \text{ kg}$

Step 2: Conversion of Work Function to Joules

To perform calculations in SI units, we convert $\Phi$ from electron-volts to Joules:

$\Phi = 1.9 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 3.0438 \times 10^{-19} \text{ J}$ [Per the definition of electron-volt energy conversion]

Step 3: Calculating Threshold Frequency ($\nu_0$)

Using the relation $\Phi = h\nu_0$ [Per Planck-Einstein relation]:

$\nu_0 = \frac{\Phi}{h} = \frac{3.0438 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J}\cdot\text{s}} = 4.594 \times 10^{14} \text{ s}^{-1}$

Step 4: Calculating Threshold Wavelength ($\lambda_0$)

Using the relation $\lambda_0 = \frac{c}{\nu_0}$:

$\lambda_0 = \frac{3 \times 10^8 \text{ m/s}}{4.594 \times 10^{14} \text{ s}^{-1}} = 6.53 \times 10^{-7} \text{ m} = 653 \text{ nm}$

Step 5: Kinetic Energy ($E_k$) of Ejected Photoelectron

Energy of incident photon ($E_{inc}$) = $\frac{hc}{\lambda}$:

$E_{inc} = \frac{(6.626 \times 10^{-34} \text{ J}\cdot\text{s})(3 \times 10^8 \text{ m/s})}{500 \times 10^{-9} \text{ m}} = 3.9756 \times 10^{-19} \text{ J}$

Using Einstein’s Photoelectric Equation: $E_k = E_{inc} - \Phi$ [Per Law of Conservation of Energy]:

$E_k = 3.9756 \times 10^{-19} \text{ J} - 3.0438 \times 10^{-19} \text{ J} = 9.318 \times 10^{-20} \text{ J}$

Step 6: Velocity ($v$) of the Photoelectron

Using $E_k = \frac{1}{2}m_ev^2$:

$v = \sqrt{\frac{2 \times E_k}{m_e}} = \sqrt{\frac{2 \times 9.318 \times 10^{-20} \text{ J}}{9.11 \times 10^{-31} \text{ kg}}}$

$v = \sqrt{2.045 \times 10^{11} \text{ m}^2/\text{s}^2} = 4.522 \times 10^5 \text{ m/s}$

Final Solution: (a) Threshold wavelength = $653 \text{ nm}$, Threshold frequency = $4.594 \times 10^{14} \text{ Hz}$; (b) Kinetic energy = $9.318 \times 10^{-20} \text{ J}$, Velocity = $4.522 \times 10^5 \text{ m/s}$.

Step 1: Establishing the Theoretical Framework

The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is governed by the Bohr Model of the atom. The relationship between the energy ($E_n$) and the principal quantum number ($n$) is defined by the following expression:

$E_n = -\frac{R_H}{n^2}$

Where:

• $E_n$ is the energy of the orbit with principal quantum number $n$.
• $R_H$ is the Rydberg constant for energy, given as $2.18 \times 10^{-18} \, \text{J atom}^{-1}$ [Per Bohr's Theory of the Hydrogen Atom].
• $n$ is the integer representing the orbit number.

Step 2: Identifying Given Variables

Based on the problem statement:

• Energy of the first orbit ($n_1 = 1$): $E_1 = -2.18 \times 10^{-18} \, \text{J atom}^{-1}$
• Target orbit: $n_2 = 5$

Step 3: Calculating Energy for the Fifth Orbit

To find the energy of the fifth orbit ($E_5$), we substitute $n = 5$ into the governing equation [By Application of Bohr's Energy Formula]:

$E_5 = -\frac{2.18 \times 10^{-18} \, \text{J}}{n^2}$

$E_5 = -\frac{2.18 \times 10^{-18}}{(5)^2} \, \text{J}$

$E_5 = -\frac{2.18 \times 10^{-18}}{25} \, \text{J}$

Step 4: Algebraic Computation

Perform the division:

$E_5 = -0.0872 \times 10^{-18} \, \text{J}$

Expressing the result in standard scientific notation [By Mathematical Convention]:

$E_5 = -8.72 \times 10^{-20} \, \text{J atom}^{-1}$

Visualization of Orbital Energy Levels

Final Solution: The energy associated with the fifth orbit of the hydrogen atom is $-8.72 \times 10^{-20} \, \text{J atom}^{-1}$.

Step 1: Evaluation of Quantum Number Constraints

To determine the validity of the set $(n = 2, l = 1, m_l = 0, m_s = - \frac{1}{2})$, we must test each quantum number against the established rules of quantum mechanics and the Pauli Exclusion Principle.

• Principal Quantum Number ($n$): Represents the shell number. $n$ must be a positive integer ($n = 1, 2, 3, \dots$). Here, $n = 2$, which is valid.
• Azimuthal Quantum Number ($l$): Represents the subshell. For a given $n$, $l$ must satisfy $0 \le l \le (n - 1)$. Here, $l = 1$. Since $1 \le (2 - 1)$ is true, this is valid.
• Magnetic Quantum Number ($m_l$): Represents the orientation of the orbital. $m_l$ must satisfy $-l \le m_l \le +l$. Given $l = 1$, the possible values for $m_l$ are $\{-1, 0, +1\}$. Since $0$ is within this range, this is valid.
• Spin Quantum Number ($m_s$): Represents the spin angular momentum of the electron. It must be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Here, $m_s = -\frac{1}{2}$, which is valid.

Step 2: Logical Synthesis of the Set

The set describes an electron residing in the $2p$ subshell. Specifically, the quantum numbers define one of the three $p$-orbitals ($2p_z$ if $m_l=0$ by convention) and specifies one of the two possible spin states for the electron.

Step 3: Theoretical Justification

Per the rules derived from the Schrödinger wave equation solutions for a hydrogenic atom [Bohr-Sommerfeld/Schrödinger theory]:

Quantum Number	Constraint Rule	Provided Value	Status
$n$	$n \in \{1, 2, 3, \dots\}$	$2$	Valid
$l$	$0 \le l \le n-1$	$1$	Valid
$m_l$	$-l \le m_l \le +l$	$0$	Valid
$m_s$	$\pm \frac{1}{2}$	$-\frac{1}{2}$	Valid

Final Solution: The set of quantum numbers $(n = 2, l = 1, m_l = 0, m_s = – \frac{1}{2})$ is a physically possible set as it satisfies all governing constraints of quantum mechanics. It corresponds to an electron located in the $2p$ subshell with spin-down configuration.

Step 1: Given Variables & Initial Setup

To determine the velocity of the photoelectron, we utilize the Photoelectric Effect equation established by Einstein: $E_{photon} = W + K.E._{max}$.

• Wavelength ($\lambda$): $4 \times 10^{-7} \, \text{m}$
• Work Function ($W$): $2.13 \, \text{eV}$
• Planck’s Constant ($h$): $6.626 \times 10^{-34} \, \text{J}\cdot\text{s}$
• Speed of Light ($c$): $3 \times 10^8 \, \text{m/s}$
• Mass of electron ($m_e$): $9.109 \times 10^{-31} \, \text{kg}$
• Conversion factor: $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$

Step 2: Calculating the Energy of the Incident Photon ($E$)

Using the Planck-Einstein relation [Per $E = \frac{hc}{\lambda}$]:

$E = \frac{(6.626 \times 10^{-34} \, \text{J}\cdot\text{s}) \times (3 \times 10^8 \, \text{m/s})}{4 \times 10^{-7} \, \text{m}}$

$E = 4.9695 \times 10^{-19} \, \text{J}$

Step 3: Converting Work Function to Joules

Convert the work function ($W$) from electron-volts to Joules to maintain dimensional consistency [Per $W(J) = W(eV) \times 1.602 \times 10^{-19}$]:

$W = 2.13 \times 1.602 \times 10^{-19} \, \text{J}$

$W = 3.4123 \times 10^{-19} \, \text{J}$

Step 4: Calculating Kinetic Energy ($K.E.$)

Subtract the work function from the incident photon energy [Per Conservation of Energy]:

$K.E. = E - W$

$K.E. = 4.9695 \times 10^{-19} - 3.4123 \times 10^{-19} \, \text{J}$

$K.E. = 1.5572 \times 10^{-19} \, \text{J}$

Step 5: Determining Velocity ($v$)

Using the classical kinetic energy formula $K.E. = \frac{1}{2} m_e v^2$, solve for $v$:

$v = \sqrt{\frac{2 \times K.E.}{m_e}}$

$v = \sqrt{\frac{2 \times 1.5572 \times 10^{-19} \, \text{kg}\cdot\text{m}^2/\text{s}^2}{9.109 \times 10^{-31} \, \text{kg}}}$

$v = \sqrt{0.3418 \times 10^{12} \, \text{m}^2/\text{s}^2}$

$v \approx 5.846 \times 10^5 \, \text{m/s}$

Final Solution: The velocity of the emitted photoelectron is approximately 5.85 \times 10^5 \, \text{m/s}.

Step 1: Theoretical Foundation

The energy ($E$) of a single photon is directly proportional to the frequency ($\nu$) of the electromagnetic radiation. This relationship is defined by Planck's Quantum Theory, expressed as:

$E = h\nu$

[Per Planck's Quantum Theory of Radiation]

Step 2: Defining Constants and Variables

To calculate the energy, we define the following parameters:

• Frequency ($\nu$) = $3 \times 10^{15} \text{ Hz}$ (or $3 \times 10^{15} \text{ s}^{-1}$)
• Planck’s Constant ($h$) = $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$

Step 3: Calculating Energy

Substitute the given values into the Planck equation:

$E = (6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (3 \times 10^{15} \text{ s}^{-1})$

Perform the algebraic multiplication of the coefficients:

$6.626 \times 3 = 19.878$

Apply the laws of exponents for the powers of 10:

$10^{-34} \times 10^{15} = 10^{-34 + 15} = 10^{-19}$

Combine the results:

$E = 19.878 \times 10^{-19} \text{ J}$

Step 4: Standard Scientific Notation

To express the result in proper scientific notation ($a \times 10^n$ where $1 \le a < 10$), adjust the decimal point:

$E = 1.9878 \times 10^1 \times 10^{-19} \text{ J}$

$E = 1.9878 \times 10^{-18} \text{ J}$

Final Solution: The energy of each photon is $1.9878 \times 10^{-18} \text{ J}$.

Step 1: Establishing Algebraic Variables

Let the symbol of the ion be represented as $_{Z}^{A}X^{n}$, where:

• $A = 37$ (Mass number given).
• $n = -1$ (Charge on the ion).
• $Z = \text{Atomic number (number of protons)}$.
• $e = \text{Number of electrons}$.
• $n_0 = \text{Number of neutrons}$.

Since the ion has a $-1$ charge, the number of electrons is related to the atomic number by the equation: $e = Z + 1$ [Per the definition of an anion].

The number of neutrons is related to the mass number by: $n_0 = A - Z = 37 - Z$ [Per the relation $A = Z + n_0$].

Step 2: Formulating the Mathematical Expression

The problem states that the ion contains 11.1% more neutrons than electrons. This is expressed mathematically as:

$n_0 = e + 0.111e$

$n_0 = 1.111e$

Step 3: Solving for Z (Atomic Number)

Substitute the expressions for $n_0$ and $e$ into the equation:

$37 - Z = 1.111(Z + 1)$

Expand the right side:

$37 - Z = 1.111Z + 1.111$

Rearrange the terms to isolate $Z$:

$37 - 1.111 = 1.111Z + Z$

$35.889 = 2.111Z$

$Z = \frac{35.889}{2.111} \approx 17$

[Calculation justification: $17 \times 2.111 \approx 35.887$, which rounds to 17 given the approximation of 11.1% as $1/9$].

Step 4: Identifying the Element and Ion

An element with atomic number $Z = 17$ is Chlorine ($Cl$).

The ion is a negatively charged ion ($Cl^-$) with a mass number of 37. Thus, the isotope is $^{37}Cl^-$.

Step 5: Structural Verification

Particle	Calculated Value
Protons ($Z$)	17
Electrons ($Z+1$)	18
Neutrons ($37-17$)	20
Check ($18 \times 1.111$)	19.998 $\approx 20$

Final Solution: The symbol of the ion is $^{37}_{17}\text{Cl}^-$.

Step 1: Identifying Given Parameters

To determine the radius of a single carbon atom from a linear arrangement, we first identify the physical properties provided in the problem statement:

• Total number of atoms ($N$): $2.0 \times 10^8$
• Total length of the arrangement ($L$): $2.4 \text{ cm}$

Step 2: Calculating the Diameter of a Single Atom

Since the atoms are arranged side by side in a single line, the total length is the sum of the diameters of all individual atoms. Assuming the atoms are spheres in contact [Geometric Assumption for linear packing]:

$L = N \times d$

Where $d$ is the diameter of one atom. Rearranging for $d$:

$d = \frac{L}{N}$

$d = \frac{2.4 \text{ cm}}{2.0 \times 10^8}$

$d = 1.2 \times 10^{-8} \text{ cm}$

Step 3: Calculating the Radius of the Atom

The radius ($r$) of a sphere is defined as half of its diameter [By definition of a circle/sphere]:

$r = \frac{d}{2}$

$r = \frac{1.2 \times 10^{-8} \text{ cm}}{2}$

$r = 0.6 \times 10^{-8} \text{ cm}$

Step 4: Unit Conversion to Scientific Notation

To express the radius in standard scientific notation:

$r = 6.0 \times 10^{-9} \text{ cm}$

Optionally, converting to picometers ($\text{pm}$) where $1 \text{ cm} = 10^{10} \text{ pm}$:

$r = 6.0 \times 10^{-9} \text{ cm} \times 10^{10} \text{ pm/cm} = 60 \text{ pm}$

Visual Representation of Linear Arrangement

Final Solution: The radius of the carbon atom is $6.0 \times 10^{-9} \text{ cm}$ or $60 \text{ pm}$.

Step 1: Analyzing the Quantum Number Constraints

To determine the validity of the set $(n = 0, l = 0, m_l = 0, m_s = + \frac{1}{2})$, we must evaluate each value against the governing principles of quantum mechanics [Per the Schrödinger wave equation and the Aufbau principle].

• Principal Quantum Number ($n$): Represents the shell or energy level of an electron. Per quantum mechanical definitions, $n$ must be a positive integer ($n = 1, 2, 3, \dots$). It cannot be zero or negative.
• Azimuthal Quantum Number ($l$): Represents the subshell (shape of the orbital). The possible values for $l$ range from $0$ to $(n - 1)$.
• Magnetic Quantum Number ($m_l$): Represents the orientation of the orbital. The possible values range from $-l$ to $+l$, including zero.
• Spin Quantum Number ($m_s$): Represents the intrinsic angular momentum (spin) of the electron, which is restricted to $\pm \frac{1}{2}$.

Step 2: Evaluating the Provided Set

Quantum Number	Provided Value	Theoretical Requirement	Status
$n$	$0$	$n \in \{1, 2, 3, \dots\}$	Invalid
$l$	$0$	$0 \leq l \leq (n-1)$	Invalid (dependent on $n$)
$m_l$	$0$	$-l \leq m_l \leq +l$	Invalid (dependent on $l$)
$m_s$	$+1/2$	$\pm 1/2$	Valid

Step 3: Reasoning and Conclusion

The set is physically impossible for the following reason: The Principal Quantum Number ($n$) defines the energy level and must satisfy the condition $n \geq 1$. Since the given value is $n = 0$, there is no corresponding shell in the atomic model where an electron can reside. Furthermore, if $n=0$, the condition $0 \leq l \leq (n-1)$ cannot be satisfied, rendering $l=0$ mathematically inconsistent with the requirement. [By the Pauli Exclusion Principle and Bohr's postulate on quantized energy levels].

Final Solution: The set of quantum numbers $(n = 0, l = 0, m_l = 0, m_s = + \frac{1}{2})$ is not possible because the principal quantum number $n$ cannot be zero; it must be a positive integer starting from 1.

Step 1: Theoretical Foundation of the Balmer Series

The Rydberg formula for the wavenumber ($\bar{\nu}$) of transitions in the hydrogen atom spectrum is defined as:

$\bar{\nu} = R_H \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

Where:

• $R_H$ is the Rydberg constant, approximately $109,677 \, \text{cm}^{-1}$.
• $n_1$ is the lower energy level. For the Balmer series, $n_1 = 2$ [Per the Bohr Model of the Hydrogen Atom].
• $n_2$ is the higher energy level.

Step 2: Identifying the Longest Wavelength Transition

Wavelength ($\lambda$) and wavenumber ($\bar{\nu}$) are inversely proportional according to the relation $\bar{\nu} = \frac{1}{\lambda}$ [Relationship between electromagnetic wave properties]. To obtain the longest wavelength, we must identify the smallest energy transition, which corresponds to the smallest possible value for $\bar{\nu}$.

For the Balmer series, the transitions occur from $n_2$ to $n_1 = 2$. The transition with the smallest energy gap occurs between the first excited state above the base level, i.e., $n_2 = 3$ to $n_1 = 2$.

Step 3: Numerical Substitution and Calculation

Substitute $n_1 = 2$ and $n_2 = 3$ into the Rydberg equation:

$\bar{\nu} = 109,677 \, \text{cm}^{-1} \times \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$

Perform the fractional arithmetic:

$\bar{\nu} = 109,677 \times \left( \frac{1}{4} - \frac{1}{9} \right)$

$\bar{\nu} = 109,677 \times \left( \frac{9 - 4}{36} \right)$

$\bar{\nu} = 109,677 \times \left( \frac{5}{36} \right)$

$\bar{\nu} = 109,677 \times 0.13888...$

$\bar{\nu} = 15,232.916 \, \text{cm}^{-1}$

Visual Representation of the Transition

Step 4: Precision Analysis

The calculation yields a wavenumber of $15,233 \, \text{cm}^{-1}$ when rounded to the nearest integer. This represents the lowest energy photon in the visible Balmer series, corresponding to the $H_\alpha$ spectral line.

Final Solution: The wavenumber for the longest wavelength transition in the Balmer series is $15,232.92 \, \text{cm}^{-1}$.

Given Variables & Initial Setup

To determine the number of photons required to provide a total energy ($E_{total}$) of $1\, \text{J}$ for light with a wavelength ($\lambda$) of $4000\, \text{pm}$, we utilize the Planck-Einstein relation.

• Total Energy ($E_{total}$) = $1\, \text{J}$
• Wavelength ($\lambda$) = $4000\, \text{pm} = 4000 \times 10^{-12}\, \text{m} = 4 \times 10^{-9}\, \text{m}$
• Planck's constant ($h$) = $6.626 \times 10^{-34}\, \text{J}\cdot\text{s}$
• Speed of light ($c$) = $3.0 \times 10^8\, \text{m/s}$

Step 1: Calculating Energy of a Single Photon ($E_{photon}$)

The energy of a single photon is defined by the equation [Per Planck’s Quantum Theory]:

$E_{photon} = \frac{hc}{\lambda}$

Substituting the given values:

$E_{photon} = \frac{(6.626 \times 10^{-34}\, \text{J}\cdot\text{s}) \times (3.0 \times 10^8\, \text{m/s})}{4 \times 10^{-9}\, \text{m}}$

$E_{photon} = \frac{19.878 \times 10^{-26}}{4 \times 10^{-9}}\, \text{J}$

$E_{photon} = 4.9695 \times 10^{-17}\, \text{J}$

Step 2: Calculating the Number of Photons ($n$)

The total energy is the product of the number of photons ($n$) and the energy per photon. Thus, $n = \frac{E_{total}}{E_{photon}}$ [By the Law of Conservation of Energy]:

$n = \frac{1\, \text{J}}{4.9695 \times 10^{-17}\, \text{J}}$

$n \approx 0.2012 \times 10^{17}$

$n \approx 2.012 \times 10^{16}$

Visual Representation of Photon Energy Quantization

Final Solution: The number of photons required to provide 1 J of energy is $2.012 \times 10^{16}$.

Step 1: Identifying the Atomic Number ($Z$)

For any neutral atom, the number of electrons is equal to the number of protons, which defines the atomic number ($Z$).

• Number of electrons ($e^-$) = $29$
• Atomic Number ($Z$) = $29$

[Per the definition of atomic structure in a neutral state, $Z = \text{number of protons} = \text{number of electrons}$].

Step 2: Applying the Aufbau Principle

To determine the electronic configuration, electrons are filled into atomic orbitals in increasing order of energy, starting from the $1s$ orbital. The sequence follows the $(n + l)$ rule, where orbitals with lower $(n + l)$ values are filled first [Per the Aufbau Principle].

The standard order of energy for subshells is: $1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p \dots$

Step 3: Distributing 29 Electrons

We distribute the 29 electrons according to the capacity of each subshell ($s=2, p=6, d=10$):

Subshell	Electrons Added	Cumulative Total
$1s$	2	2
$2s$	2	4
$2p$	6	10
$3s$	2	12
$3p$	6	18
$4s$	1	19
$3d$	10	29

Step 4: Theoretical Justification for Anomalous Configuration

While the expected configuration based on the Aufbau principle is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^9$, the actual configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 3d^{10}$.

Reasoning: Half-filled and fully-filled $d$-orbitals provide extra stability due to symmetry and increased exchange energy [Per the Hund’s Rule and Stability of Half-filled/Fully-filled Subshells]. Moving one electron from $4s$ to $3d$ completes the $d$-subshell ($3d^{10}$), resulting in a more stable ground state.

Final Configuration Summary

The electronic configuration of the element (Copper, $Cu$) is:

Final Solution: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$ (or $[Ar] 3d^{10} 4s^1$)

Step 1: Evaluation of the Provided Quantum Number Set

The given set of quantum numbers is: $n = 2$, $l = 1$, $m_l = 0$, $m_s = -\frac{1}{2}$. To determine the validity of this set, we must evaluate each parameter against the established rules of quantum mechanics governing atomic orbitals.

Step 2: Verification Against Quantum Mechanical Rules

• Principal Quantum Number ($n$): $n = 2$. This value must be a positive integer ($n = 1, 2, 3, \dots$). $n = 2$ is valid [Per the Bohr-Sommerfeld model and Schrödinger wave equation].
• Azimuthal Quantum Number ($l$): $l = 1$. The rule states that for a given $n$, $l$ can range from $0$ to $n-1$. Here, $n-1 = 2-1 = 1$. Since $0 \leq 1 \leq 1$, this value is valid [Per the definition of angular momentum quantization].
• Magnetic Quantum Number ($m_l$): $m_l = 0$. The rule states that for a given $l$, $m_l$ can range from $-l$ to $+l$, including zero. Given $l = 1$, the allowed values are $-1, 0, +1$. Since $0$ is within this range, this value is valid [Per the spatial orientation quantization rule].
• Spin Quantum Number ($m_s$): $m_s = -\frac{1}{2}$. The rule states that for any orbital, the electron spin can only be $+\frac{1}{2}$ or $-\frac{1}{2}$. This value is valid [Per the Pauli Exclusion Principle and the Stern-Gerlach experiment].

Step 3: Orbital Assignment and Logical Synthesis

The set describes an electron in the $2p$ subshell. Specifically, since $n=2$ and $l=1$, the orbital is a $2p$ orbital. The magnetic quantum number $m_l = 0$ designates the $2p_z$ orbital (by convention). The spin $m_s = -\frac{1}{2}$ indicates a specific spin orientation.

Step 4: Final Conclusion

Since the provided set ($n=2, l=1, m_l=0, m_s=-\frac{1}{2}$) satisfies all constraints imposed by the rules of quantum mechanics, it represents a physically possible state for an electron within an atom.

Final Solution: The set of quantum numbers $n = 2$, $l = 1$, $m_l = 0$, $m_s = -\frac{1}{2}$ is possible. It corresponds to an electron located in a $2p$ orbital.

Step 1: Establishing Physical Constants and Variables

To calculate the frequency ($\nu$) and energy ($E$), we utilize the relationship between the speed of light ($c$), wavelength ($\lambda$), and Planck's constant ($h$).

• Speed of light ($c$) = $2.9979 \times 10^8 \text{ m/s}$
• Planck's constant ($h$) = $6.626 \times 10^{-34} \text{ J}\cdot\text{s}$
• Wavelength 1 ($\lambda_1$) = $589 \text{ nm} = 589 \times 10^{-9} \text{ m}$
• Wavelength 2 ($\lambda_2$) = $589.6 \text{ nm} = 589.6 \times 10^{-9} \text{ m}$

Step 2: Calculating Frequencies for Each Transition

Using the fundamental relation $\nu = \frac{c}{\lambda}$ [Per the Wave-Particle Duality relationship]:

For $\lambda_1 = 589 \times 10^{-9} \text{ m}$:

$\nu_1 = \frac{2.9979 \times 10^8 \text{ m/s}}{589 \times 10^{-9} \text{ m}} = 5.0898 \times 10^{14} \text{ s}^{-1}$

For $\lambda_2 = 589.6 \times 10^{-9} \text{ m}$:

$\nu_2 = \frac{2.9979 \times 10^8 \text{ m/s}}{589.6 \times 10^{-9} \text{ m}} = 5.0846 \times 10^{14} \text{ s}^{-1}$

Step 3: Calculating Energy Difference Between Excited States

The energy of a photon is given by $E = h\nu$ [Per Planck’s Quantum Theory]. The difference in energy ($\Delta E$) between the two states corresponding to the transitions is expressed as:

$\Delta E = E_1 - E_2 = h\nu_1 - h\nu_2 = h(\nu_1 - \nu_2)$

Calculate the frequency difference ($\Delta\nu$):

$\Delta\nu = (5.0898 \times 10^{14}) - (5.0846 \times 10^{14}) = 0.0052 \times 10^{14} \text{ s}^{-1} = 5.2 \times 10^{11} \text{ s}^{-1}$

Calculate the energy difference ($\Delta E$):

$\Delta E = (6.626 \times 10^{-34} \text{ J}\cdot\text{s}) \times (5.2 \times 10^{11} \text{ s}^{-1})$ $\Delta E = 3.4455 \times 10^{-22} \text{ J}$

Visual Representation of Atomic Absorption

Transition	Frequency ($\text{s}^{-1}$)
$\lambda_1$	$5.09 \times 10^{14}$
$\lambda_2$	$5.08 \times 10^{14}$

Final Solution: The frequencies are $5.09 \times 10^{14} \text{ s}^{-1}$ and $5.08 \times 10^{14} \text{ s}^{-1}$ respectively, and the energy difference between the excited states is $3.45 \times 10^{-22} \text{ J}$.

Step 1: Identification of Atomic Properties

The oxygen atom ($\text{O}$) possesses an atomic number ($Z$) of 8. This indicates that a neutral oxygen atom contains 8 protons and, consequently, 8 electrons in its ground state [Per the definition of atomic number in the Bohr-Rutherford model].

Step 2: Accounting for Ionic Charge

The ion in question is $\text{O}^{2-}$, which is an oxide anion. The $2-$ charge indicates the gain of two additional electrons beyond the ground state configuration of the neutral atom. [Per the Octet Rule, atoms gain or lose electrons to achieve a stable valence shell configuration similar to noble gases].

Total number of electrons in $\text{O}^{2-} = Z + (\text{magnitude of charge}) = 8 + 2 = 10$ electrons.

Step 3: Applying Aufbau Principle and Hund's Rule

To determine the electronic configuration, we distribute 10 electrons according to the energy hierarchy: $1s < 2s < 2p$. [Per the Aufbau Principle, electrons fill orbitals in order of increasing energy].

• Subshell 1s: Capacity of 2 electrons. Occupied: $1s^2$.
• Subshell 2s: Capacity of 2 electrons. Occupied: $2s^2$.
• Subshell 2p: Capacity of 6 electrons. Remaining electrons: $10 - 2 - 2 = 6$. Occupied: $2p^6$.

Step 4: Visualization of Orbital Occupancy

Step 5: Tabular Summary

Shell/Subshell	Electronic Capacity	Occupancy in $\text{O}^{2-}$
$1s$	2	2
$2s$	2	2
$2p$	6	6

Final Solution: The electronic configuration of the oxide ion ($\text{O}^{2-}$) is $1s^2 2s^2 2p^6$.

Step 1: Identifying the Physical Principles

The energy $E$ of a single photon is directly proportional to the frequency $\nu$ and inversely proportional to the wavelength $\lambda$. The relationship is governed by the Planck-Einstein relation [Per the Planck-Einstein Relation]:

$E = \frac{hc}{\lambda}$

Where:

• $h$ (Planck’s constant) = $6.626 \times 10^{-34} \text{ J s}$
• $c$ (Speed of light) = $3.0 \times 10^8 \text{ m s}^{-1}$
• $\lambda$ (Wavelength) = $242 \text{ nm} = 242 \times 10^{-9} \text{ m}$

Step 2: Calculating Energy per Photon (J)

Substituting the given values into the equation:

$E = \frac{(6.626 \times 10^{-34} \text{ J s}) \times (3.0 \times 10^8 \text{ m s}^{-1})}{242 \times 10^{-9} \text{ m}}$

$E = \frac{1.9878 \times 10^{-25}}{242 \times 10^{-9}} \text{ J}$

$E = 8.214 \times 10^{-19} \text{ J per atom}$

Step 3: Calculating Ionization Energy per Mole

To find the ionization energy for one mole of sodium atoms, we multiply the energy of a single photon by Avogadro’s constant ($N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$) [Per the Mole Concept Definition]:

$E_{\text{mole}} = E \times N_A$

$E_{\text{mole}} = (8.214 \times 10^{-19} \text{ J atom}^{-1}) \times (6.022 \times 10^{23} \text{ atom mol}^{-1})$

$E_{\text{mole}} = 4.946 \times 10^5 \text{ J mol}^{-1}$

Step 4: Unit Conversion to kJ mol$^{-1}$

Convert Joules to kilojoules by dividing by $10^3$ [Per Standard SI Conversion Factors]:

$E_{\text{kJ}} = \frac{4.946 \times 10^5}{10^3} \text{ kJ mol}^{-1}$

$E_{\text{kJ}} = 494.6 \text{ kJ mol}^{-1}$

Final Solution: The ionization energy of sodium is 494.6 kJ mol$^{-1}$.