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Structure of Atom appeared 39 times across 3 years — 4.5% of Chemistry. This question is from Bohr Model of Atom.

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Which of the following postulate of Bohr's model of hydrogen atom is not in agreement with quantum mechanical model of an atom?

Solution & Explanation

Related Formula
Eₙ = -(13.6 · Z²)/(n²) eV Δ E = E₂ - E₁ = hν
Core Logic

According to quantum mechanics (wave mechanics and Heisenberg's uncertainty principle), electrons do not move in fixed two-dimensional circular orbits around the nucleus. Instead, their positions are represented by three-dimensional probability density functions called orbitals.

Step 1: Evaluation of Options

Statements 1, 2, and 3 are consistent with both models (quantized energy states and non-radiating stationary states).

Statement 4 asserts that electrons move in well-defined circular paths, which violates quantum mechanical principles where electron motion is three-dimensional and probabilistic.

Pattern Recognition

Bohr model = 2D planar circular orbits. Quantum Mechanical model = 3D spatial electron clouds (orbitals). Fixed circular path is strictly a Bohr concept not supported by quantum mechanics.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Reference Study Guides

Structure Of Atom

More Structure of Atom Previous-Year Questions — Page 8

Q90 jee_main_2024_30_january_evening Atomic Spectra
Number of spectral lines obtained in He^+ spectra, when an electron makes transition from fifth excited state to first excited state will be
Numerical Answer. Answer: 10 to 10

Solution

Related Formula
Number of spectral lines = (Δ n (Δ n + 1))/(2)

where Δ n = n₂ - n₁

Core Logic

Identify the principal quantum numbers (n) for the given states:

  • Fifth excited state means n₂ = 5 + 1 = 6.
  • First excited state means n₁ = 1 + 1 = 2.

    • Calculate the difference:

Δ n = n₂ - n₁ = 6 - 2 = 4
Step 1: Calculate Total Spectral Lines

Substitute Δ n = 4 into the formula:

Maximum number of spectral lines = (4(4 + 1))/(2) = (4 × 5)/(2)

= 10

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q68 jee_main_2024_30_jan_morning Quantum Mechanical Model
Given below are two statements: Statement-I: The orbitals having same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given
  • A. Statement-I is true but Statement-II is false
  • B. Both Statement-I and Statement-II are true.
  • C. Both Statement-I and Statement-II are false
  • D. Statement-I is false but Statement-II is true

Solution

Core Logic

Statement-I defines degenerate orbitals correctly: Orbitals that share the exact same energy level are called degenerate orbitals. Statement-II claims 3p and 3d are not degenerate in a hydrogen atom. For a single-electron system like Hydrogen (1s¹), the energy of an orbital depends only on the principal quantum number 'n'.

Step 1: Final conclusion

In a hydrogen atom, energy of 3s = 3p = 3d. Therefore, 3p and 3d are degenerate. Thus, Statement-II is false.

Pattern Recognition

Hydrogen atom (single electron species) = Energy dictated strictly by n. Multi-electron atoms = Energy dictated by (n + l) rule.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q jee_main_2024_31_jan_evening Quantum Numbers
The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are
  • A. n = 4, l = 2, m = -1, s = +(1)/(2)
  • B. n = 4, l = 0, m = 0, s = +(1)/(2)
  • C. n = 3, l = 0, m = 1, s = +(1)/(2)
  • D. n = 2, l = 0, m = 0, s = +(1)/(2)

Solution

Core Logic

The atomic number of potassium (K) is 19. Its electronic configuration is: 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹. The outermost orbital of potassium is the 4s orbital.

For the 4s orbital: Principal quantum number, n = 4 Azimuthal quantum number, l = 0 (for s-subshell) Magnetic quantum number, m = 0 (since l = 0) Spin quantum number, s = ± (1)/(2)

Step 1: Final Matching

Comparing these values with the options, option (2) provides n = 4, l = 0, m = 0, s = +(1)/(2), which perfectly describes the outermost electron.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q89 jee_main_2024_31_jan_morning Photoelectric Effect and Ionization Energy
The ionization energy of sodium in kJ mol⁻¹. If electromagnetic radiation of wavelength 242 nm is just sufficient to ionize sodium atom is ________
Numerical Answer. Answer: 493 to 495

Solution

Related Formula
E = 1240λ (nm) eV Etotal = E × NA
Step 1: Energy per atom
E = (1240)/(242) eV = 5.12 eV

Converting eV to Joules:

E = 5.12 × 1.6 × 10⁻¹⁹ J/atom E = 8.192 × 10⁻¹⁹ J/atom
Step 2: Energy per mole

To find the ionization energy per mole, multiply by Avogadro's number (NA ≈ 6.022 × 10²³):

IE = 8.192 × 10⁻¹⁹ × 6.022 × 10²³ J/mol IE ≈ 493.3 × 10³ J/mol ≈ 494 kJ/mol
Chapter Mix

Class 11 Chemistry: Structure of Atom

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