JEE Main · Physics ↓ Falling

Atoms appeared 22 times across 3 years — 2.5% of Physics. This question is from Alpha-particle Scattering and Rutherford's Nuclear Model.

Year 2026 2025 2024 Total
Questions 5 8 9 22

Given below are two statements: Statement I: Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model. Statement II: An atom is a spherical cloud of positive charges with electrons embedded in it, is a special case of Rutherford's model. In the light of the above statements, choose the most appropriate from the options given below:

Solution & Explanation

Core Logic
  • Statement I: According to Rutherford's planetary model of the atom, almost the entire mass of an atom and all of its positive charge are concentrated in a centrally located small volume called the nucleus, with electrons revolving around it. Thus, Statement I is true.
  • Statement II: Thomson's model of the atom (plum-pudding model) describes the atom as a spherical cloud of positive charge with electrons embedded in it. This model was proposed before Rutherford's nuclear model and is not a special case of Rutherford's model. Thus, Statement II is false.
Step 1: Conclusion

Therefore, Statement I is true but Statement II is false.

Pattern Recognition

Understand the evolutionary timeline of atomic models:

  • Thomson's Plum Pudding Model (spherical cloud with embedded electrons)
  • Rutherford's Nuclear Model (planetary orbital model with tiny centralized nucleus)
  • They are conceptually distinct, making Statement II false.

Chapter Mix

Class 12 Physics: Atoms

Reference Study Guides

More Atoms Previous-Year Questions — Page 5

Q53 jee_main_2024_30_jan_morning Hydrogen Energy Levels and Transitions
A electron of hydrogen atom on an excited state is having energy Eₙ = -0.85 ~eV. The maximum number of allowed transitions to lower energy level is ....
Numerical Answer. Answer: 6 to 6

Solution

Related Formula
Eₙ = -(13.6)/(n²) ~eV Number of transitions = (n(n - 1))/(2)
Core Logic

First, identify the principal quantum number n corresponding to the energy -0.85 ~eV. Then, use the combinatorics formula to find the total possible downward emission transitions.

Step 1: Find Quantum State
Eₙ = -(13.6)/(n²) -0.85 = -(13.6)/(n²) n² = (13.6)/(0.85) = 16

n = 4

Step 2: Calculate Transitions

Maximum number of transitions from n=4 to lower levels (n=3, 2, 1):

= (n(n - 1))/(2) = (4(4 - 1))/(2) = (12)/(2) = 6
Pattern Recognition

Energy states in Hydrogen are heavily standardized: n=1 arrow -13.6, n=2 arrow -3.4, n=3 arrow -1.51, n=4 arrow -0.85. Recognize -0.85 ~eV as state 4 immediately.

Chapter Mix

Class 12 Physics: Atoms

Q36 jee_main_2024_31_jan_morning Hydrogen Spectrum
If the wavelength of the first member of Lyman series of hydrogen is λ. The wavelength of the second member will be
  • A. (27)/(32) λ
  • B. (32)/(27)λ
  • C. (27)/(5) λ
  • D. (5)/(27)λ

Solution

Related Formula
(1)/(λ) = R Z² [ (1)/(n₁²) - (1)/(n₂²) ]
Core Logic

For the first member of the Lyman series of hydrogen (n₁ = 1, n₂ = 2):

(1)/(λ) = (13.6 Z²)/(hc) [ (1)/(1²) - (1)/(2²) ] (1)/(λ) = (13.6 Z²)/(hc) [ (3)/(4) ] (i)
Step 2: Second Member Calculation

For the second member of the Lyman series (n₁ = 1, n₂ = 3):

(1)/(λ') = (13.6 Z²)/(hc) [ (1)/(1²) - (1)/(3²) ] (1)/(λ') = (13.6 Z²)/(hc) [ (8)/(9) ] (ii)
Step 3: Ratio

On dividing equation (i) by (ii):

(λ')/(λ) = (3/4)/(8/9) = (3)/(4) × (9)/(8) (λ')/(λ) = (27)/(32) λ' = (27)/(32) λ
Pattern Recognition

Rydberg ratios between members of the same series are purely derived from the bracket terms [1/n₁² - 1/n₂²]. For Lyman 1st and 2nd, the ratio is (3/4) / (8/9) = 27/32.

Chapter Mix

Class 12 Physics: Atoms

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