CUET UG Physics Modern Physics Bohrs Model of Hydrogen Atom Energy Levels Spectral Series

Must‑Know (15–20 detailed bullets)

• Bohr’s model applies only to hydrogen and hydrogen-like atoms (e.g., He⁺, Li²⁺).
• Electrons revolve in stable orbits without radiating energy — called stationary orbits.
• Angular momentum is quantized: ( L = mvr = \frac{nh}{2\pi} ), where ( n = 1,2,3,... )
• The radius of the nth orbit in hydrogen: ( r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = 0.529 \times n^2 ) Å. For n=1, r₁ = 0.529 Å (Bohr radius).
• Total energy of electron in nth orbit: ( E_n = -\frac{13.6}{n^2} ) eV for hydrogen.
• Energy of electron in ground state (n=1) of hydrogen is –13.6 eV.
• Energy of electron in first excited state (n=2) is –3.4 eV.
• Ionization energy of hydrogen atom is +13.6 eV — energy required to remove electron from n=1 to n=∞.
• Energy difference between two levels: ( \Delta E = E_2 - E_1 = h\nu = \frac{hc}{\lambda} )
• Spectral lines arise when electron jumps from higher to lower energy level, emitting a photon of specific frequency.
• Lyman series: transitions to n=1; lies in ultraviolet region.
• Balmer series: transitions to n=2; lies in visible region; first line (n=3→2) is Hα at 656.3 nm.
• Paschen series: transitions to n=3; lies in infrared region.
• Brackett series: transitions to n=4; infrared.
• Pfund series: transitions to n=5; infrared.
• Wavelength of emitted radiation given by Rydberg formula: ( \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) ), where ( R = 1.097 \times 10^7 \, \text{m}^{-1} ) (Rydberg constant).
• For Balmer series, longest wavelength corresponds to n=3→2; shortest (series limit) is n=∞→2.
• In hydrogen, number of spectral lines emitted when electron drops from n₂ to n₁: ( \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2} ). Example: from n=4 to n=1 → 6 lines.
• Bohr model fails to explain fine structure of spectral lines and Zeeman effect.
• verify from NCERT: Exact derivation steps of radius and energy expressions using Coulomb’s law and quantization condition.

Difficulty Level

Intermediate — requires understanding of quantization, energy transitions, and application of formulas, but avoids complex derivations.

Common CUET Traps

• Trap: Assuming Balmer series includes UV lines.
  Avoid: Balmer series is only in visible and near UV; Lyman is fully UV.

• Trap: Using ( E_n = -\frac{13.6}{n} ) instead of ( E_n = -\frac{13.6}{n^2} ) eV.
  Avoid: Remember energy varies as inverse square of quantum number.

• Trap: Thinking all hydrogen spectral series are visible.
  Avoid: Only Balmer series is partially visible; others are UV or IR.

Practice MCQs

Q1. What is the energy of an electron in the second excited state of hydrogen atom?
A. –1.51 eV
B. –3.4 eV
C. –0.85 eV
D. –13.6 eV
Answer: A
Explanation: Second excited state is n=3; ( E_3 = -\frac{13.6}{9} = -1.51 ) eV.
Why others fail: B is energy for first excited state (n=2), a common misidentification.

Q2. Which spectral series of hydrogen lies entirely in the ultraviolet region?
A. Balmer
B. Paschen
C. Lyman
D. Brackett
Answer: C
Explanation: Lyman series involves transitions to n=1 and lies in UV.
Why others fail: Balmer is visible, others are IR — confusion arises due to overlapping regions.

Q3. The radius of the first Bohr orbit in He⁺ ion is approximately:
A. 0.2645 Å
B. 0.529 Å
C. 1.058 Å
D. 2.116 Å
Answer: A
Explanation: ( r = \frac{0.529 n^2}{Z} ); for He⁺, Z=2, n=1 → ( r = \frac{0.529}{2} = 0.2645 ) Å.
Why others fail: Option B is radius for H atom (Z=1), often selected if Z is ignored.

Q4. If an electron jumps from n=4 to n=2 in hydrogen atom, how many spectral lines are possible?
A. 1
B. 3
C. 6
D. 4
Answer: A
Explanation: A single electron transition from n=4 to n=2 produces one spectral line.
Why others fail: C is total lines if electron de-excites stepwise from n=4 to ground, not direct jump.

Q5. The shortest wavelength in the Balmer series of hydrogen corresponds to transition from:
A. n = 2 → n = 1
B. n = ∞ → n = 2
C. n = 3 → n = 2
D. n = ∞ → n = 1
Answer: B
Explanation: Series limit (shortest λ) in Balmer series is n=∞ to n=2.
Why others fail: D gives Lyman series limit, which is shorter but not part of Balmer.

Last‑Minute Revision

• ⚠️ Bohr radius = 0.529 Å — radius of n=1 orbit in H atom.
• ⚠️ Energy in nth orbit: ( E_n = -\frac{13.6}{n^2} ) eV.
• ⚠️ Ionization energy of H = 13.6 eV.
• ⚠️ Lyman: to n=1 (UV); Balmer: to n=2 (visible); Paschen: to n=3 (IR).
• ⚠️ Hα line: red, 656.3 nm, transition n=3→2.
• ⚠️ Rydberg constant R = 1.097 × 10⁷ m⁻¹.
• ⚠️ Angular momentum: ( mvr = \frac{nh}{2\pi} ).
• ⚠️ verify from NCERT: Formula for radius: ( r_n \propto \frac{n^2}{Z} ).
• ⚠️ verify from NCERT: Energy ( E_n \propto -\frac{Z^2}{n^2} ).
• ⚠️ For He⁺, Z=2 → energy levels 4× deeper than H.
• ⚠️ Number of spectral lines from n to ground: ( \frac{n(n-1)}{2} ).
• ⚠️ Longest λ in Balmer: n=3→2; shortest: n=∞→2.
• ⚠️ Bohr model valid only for one-electron systems.
• ⚠️ Electron does not radiate energy in stationary orbits.
• ⚠️ Mnemonic: Last Ball Pitched Brings Pitches → Lyman (UV), Balmer (Vis), Paschen, Brackett, Pfund (all IR).