Atomic Spectra Past Papers

Concept: The hydrogen spectrum is organized into distinctive series based on the final orbital (\( p \)) that an electron transitions to. The visible light region is narrow (400-700 nm).

Formula:
$$ \lambda = \frac{hc}{\Delta E} $$

Solution:

• The Balmer series is strictly defined by electrons falling from higher orbits down to the second orbit (\( p = 2 \)).
• The energy gaps for these specific transitions (e.g., \( n=3 \to 2 \), \( n=4 \to 2 \)) perfectly correspond to photons with wavelengths in the 400 nm to 700 nm range.
• This range is exclusively the Visible region of light for humans.

Why other options are incorrect:
Lyman (C) drops to \( p=1 \), releasing massive energy as UV radiation. Paschen (D) and Brackett (B) drop to \( p=3 \) and \( p=4 \), releasing smaller energy amounts as invisible Infrared radiation.

Concept: The Lyman series involves electron transitions terminating at the absolute ground state (\( n=1 \)) of the hydrogen atom, producing the most energetic spectral lines.

Formula:
$$ E_{\text{photon}} = -13.6 \text{ eV} \left( \frac{1}{n^2} - \frac{1}{1^2} \right) $$

Solution:

• Since the transitions terminate at the deeply bound \( n=1 \) shell, the energy difference \( \Delta E \) is exceptionally large (between 10.2 eV and 13.6 eV).
• Photons with these energy levels correspond to wavelengths ranging from roughly 91 nm to 121 nm.
• This wavelength span sits firmly in the Ultraviolet (UV) spectrum, completely invisible to the human eye.

Why other options are incorrect:
Visible light (D) requires less energy (Balmer series). Infrared (B) and Far-infrared (A) require even less energy transitions, corresponding to series terminating at higher shells like Paschen or Brackett.

Concept: In atomic physics, giving an atom enough energy to raise its electron from the ground state to a higher (but still bound) energy state is called "excitation."

Formula:
$$ E = qV_{\text{excitation}} = E_{\text{higher}} - E_{\text{ground}} $$

Solution:

• When a free electron is accelerated through an electric potential \( V \), it gains kinetic energy.
• If it collides with an atom and transfers this energy perfectly to lift the atom's bound electron to a higher shell, this specific voltage is defined as the "Excitation Potential."
• The electron is not completely removed from the atom; it merely jumps to a higher orbit.

Why other options are incorrect:
Ionization potential (A) is specifically the potential required to completely strip an electron from the atom (sending it to infinity), not just raise it to a higher shell. String potential (C) is a nonsense distractor term. Acceleration potential (D) is too generic and doesn't specify the bound-state atomic interaction.

Concept: Niels Bohr derived the radius of atomic orbits by combining classical mechanics (centripetal force matching electrostatic force) with his quantum postulate regarding angular momentum.

Formula:
$$ r_n = \frac{n^2 h^2}{4\pi^2 k m e^2} $$

Solution:

• In the formula above, all terms except \( n \) (Planck's constant \( h \), Coulomb's constant \( k \), electron mass \( m \), and elementary charge \( e \)) are fundamental constants for a given atom.
• Grouping these constants together yields: \( r_n = (\text{constant}) \times n^2 \).
• For hydrogen, this simplifies to \( r_n = 0.529\text{\AA} \times n^2 \).
• Therefore, the radius scales quadratically; it is strictly proportional to \( n^2 \).

Why other options are incorrect:
Options A, B, and D misrepresent the algebraic derivation. If the radius was simply proportional to \( n \), the quantum spacing of atoms would be strictly linear rather than expanding vastly outward.

Concept: Photon wavelength is inversely proportional to the energy of the atomic transition. Longest wavelength equates to the lowest possible energy jump between shells.

Formula:
$$ \Delta E = E_{\text{upper}} - E_{\text{lower}} = \frac{hc}{\lambda} $$

Solution:

• The inner shells (like K, \( n=1 \)) have massive binding energies. Transitions ending at the K-shell release large amounts of energy (short wavelength).
• The M-shell (\( n=3 \)) is much further from the nucleus. An \( M_\alpha \) photon is generated by an electron falling from the N-shell (\( n=4 \)) to the M-shell.
• Because higher orbital energy levels are spaced very closely together, the energy difference between N and M is very small.
• This incredibly small energy gap generates the lowest energy photon among the options, and therefore the longest wavelength.

Why other options are incorrect:
Options A, B, and C all represent electrons plunging deep into the intensely bound K-shell, releasing highly energetic, short-wavelength X-rays.

Concept: In the hydrogen atom spectrum, the spectral series are strictly classified by the region of the electromagnetic spectrum in which their emitted photons fall, determined by the final resting shell (\( p \)).

Formula:
$$ p=1 \rightarrow \text{UV (Lyman)} $$
$$ p=2 \rightarrow \text{Visible (Balmer)} $$
$$ p \ge 3 \rightarrow \text{IR (Paschen, Brackett, Pfund)} $$

Solution:

• Lyman is Ultraviolet (UV). Balmer is Visible.
• Paschen (\( p=3 \)), Brackett (\( p=4 \)), and Pfund (\( p=5 \)) all feature low-energy transitions that fall squarely into the Infrared (IR) region.
• Option B correctly pairs two IR series. Option C also correctly pairs two IR series.
• Therefore, Option D accurately encapsulates all correct groupings.

Why other options are incorrect:
Option A is entirely incorrect as it includes UV and Visible series. Options B and C are technically true but functionally incomplete individually compared to the overarching Option D.

Concept: Ultraviolet radiation comprises high-energy photons. In the hydrogen atom, the highest energy emissions occur when an electron drops back down to the lowest possible energy state, the ground level.

Formula:
$$ \Delta E_{\text{max}} = E_n - E_1 \quad (\text{Lyman series}) $$

Solution:

• The Lyman series is defined by electronic transitions terminating at the principle quantum shell \( p = 1 \).
• These jumps (from \( n=2,3,4\dots \) down to \( p=1 \)) cross the largest energy gaps in the hydrogen atom.
• Because \( E = hf \), large energy drops emit high-frequency, high-energy photons, which correspond exactly to the Ultraviolet (UV) spectrum.

Why other options are incorrect:
Balmer (B) falls in the visible range. Paschen (A) and Pfund (D) feature smaller energy transitions terminating at higher energy levels, placing them in the low-energy Infrared region.

Concept: According to Bohr's postulates and the principle of conservation of energy, the energy of an emitted photon exactly equals the energy difference between the initial and final energy states of the atom.

Formula:
$$ \Delta E = E_2 - E_1 $$
$$ E_{\text{photon}} = \frac{hc}{\lambda} $$

Solution:

• Set the transition energy equal to the photon energy: \( E_2 - E_1 = \frac{hc}{\lambda} \).
• To isolate wavelength (\( \lambda \)), multiply both sides by \( \lambda \) and divide by the energy difference \( (E_2 - E_1) \).
• This algebraically yields: \( \lambda = \frac{hc}{E_2 - E_1} \).

Why other options are incorrect:
Option A is the wave number (inverse wavelength \( 1/\lambda \)). Option C subtracts the higher energy from the lower energy, incorrectly producing a negative wavelength. Option D incorrectly arranges the fundamental constants \( h \) and \( c \).

Concept: The hydrogen emission spectrum is divided into specific series based on the final energy level (\( p \)) of the electron transition. Each series corresponds to a distinct region of the electromagnetic spectrum.

Formula:
$$ \text{Lyman } (p=1) \rightarrow \text{Ultraviolet} $$
$$ \text{Balmer } (p=2) \rightarrow \text{Visible} $$
$$ \text{Paschen } (p=3) \rightarrow \text{Infrared} $$

Solution:

• The Lyman series occurs when electrons transition from higher energy levels (\( n \ge 2 \)) down to the ground state (\( p = 1 \)).
• These transitions release the highest amount of energy, corresponding to the shortest wavelengths, which fall strictly in the ultraviolet region.

Why other options are incorrect:
Balmer (A) lies in the visible region. Paschen (B) and Brackett (D) both lie in the infrared region.

Concept: In characteristic X-ray spectra, the wavelength of the emitted photon is inversely proportional to the energy difference between the transitioning electron shells.

Formula:
$$ E = \frac{hc}{\lambda} \implies \lambda \propto \frac{1}{\Delta E} $$

Solution:

• A longer wavelength requires a smaller energy transition (\( \Delta E \)).
• \( K \)-series X-rays result from transitions down to the innermost K shell (\( n=1 \)), yielding massive energy drops and very short wavelengths.
• The \( M_\alpha \) line results from a transition from the N shell (\( n=4 \)) to the M shell (\( n=3 \)).
• Because energy levels get much closer together further from the nucleus, the energy difference between N and M is much smaller than any transition ending at K.
• Smallest \( \Delta E \) results in the longest wavelength \( \lambda \).

Why other options are incorrect:
Options A, B, and C all terminate at the deeply bound K shell, resulting in high-energy, extremely short-wavelength X-ray photons.

Concept: The wavelength of an emitted photon is inversely proportional to the energy transition. The shortest possible wavelength across all series corresponds to the maximum possible energy transition in the hydrogen atom.

Formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$

Solution:

• To minimize \( \lambda \), the expression \( \left( \frac{1}{p^2} - \frac{1}{n^2} \right) \) must be maximized.
• This happens when the electron falls to the lowest possible ground state (\( p = 1 \), the Lyman series) from the highest possible initial state (\( n = \infty \)).
• This maximum energy gap (13.6 eV) translates to the absolute shortest wavelength in the entire hydrogen spectrum.

Why other options are incorrect:
Balmer, Paschen, and Brackett series end at higher energy levels (\( p=2, 3, 4 \)), meaning their largest possible energy drops are always smaller than the Lyman series, resulting in strictly longer wavelengths.

Concept: The longest wavelength in any spectral series corresponds to the minimum energy transition. For the Balmer series, this occurs when an electron transitions from \( n = 3 \) to \( p = 2 \).

Formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$

Solution:

• Substitute \( p = 2 \) and \( n = 3 \) into the Rydberg equation:
• \( \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \)
• \( \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{9 - 4}{36} \right) = R_H \left( \frac{5}{36} \right) \)
• Invert the result to solve for wavelength: \( \lambda = \frac{36}{5R_H} \).

Why other options are incorrect:
The other options result from incorrect fractional subtractions or choosing the wrong series transitions.

Concept: The shortest wavelength in the Lyman series corresponds to the transition with the maximum energy. This happens when an electron falls from infinity (\( n = \infty \)) down to the ground state (\( p = 1 \)).

Formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$

Solution:

• Set \( p = 1 \) (Lyman series) and \( n = \infty \) (shortest wavelength limit).
• Substitute into the Rydberg formula: \( \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \).
• Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies strictly to \( \frac{1}{\lambda} = R_H(1 - 0) = R_H \).
• Inverting both sides gives \( \lambda = \frac{1}{R_H} \).

Why other options are incorrect:
Option A is the wave number (\( 1/\lambda \)), not the wavelength. Options C and D represent calculations for entirely different transitions or incorrect algebraic inversions.

Concept: The 'first line' of a series is the lowest energy transition. For the Lyman series (ground state \( p=1 \)), the first line corresponds to an electron falling from the closest upper level, which is \( n=2 \).

Formula:
$$ \frac{1}{\lambda} = R_H \left( \frac{1}{p^2} - \frac{1}{n^2} \right) $$

Solution:

• Substitute \( p=1 \) and \( n=2 \).
• \( \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \)
• \( \frac{1}{\lambda} = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \)
• To find \( \lambda \), invert the fraction: \( \lambda = \frac{4}{3R_H} \).

Why other options are incorrect:
Option A is an incorrect inversion leaving \( R_H \) in the numerator. Options B and C represent random algebraic errors or incorrect \( n \) values.

Concept: In the Bohr model, the total energy of an electron in a given orbit is the sum of its kinetic and potential energies. By the Virial theorem for Coulombic forces, Kinetic Energy (\( K.E. \)) is exactly equal to the negative of the Total Energy (\( T.E. \)).

Formula:
$$ K.E. = - T.E. $$

Solution:

• The total energy of an electron in the first Bohr orbit (ground state, \( n=1 \)) of hydrogen is \( -13.6 \text{ eV} \).
• Kinetic energy is a scalar quantity of motion and must always be positive.
• Therefore, \( K.E. = -(-13.6 \text{ eV}) = +13.6 \text{ eV} \).

Why other options are incorrect:
Option C (\( -13.6 \text{ eV} \)) represents the Total Energy, not kinetic energy. Option A uses the wrong unit (Joules instead of electron-volts). Option D uses incorrect scaling (keV).