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Atoms appeared 22 times across 3 years — 2.5% of Physics. This question is from Bohr Model of the Hydrogen Atom.

Year 2026 2025 2024 Total
Questions 5 8 9 22

During the transition of electron from state A to state C of a Bohr atom, the wavelength of emitted radiation is 2000 Å and it becomes 6000 Å when the electron jumps from state B to state C. Then the wavelength of the radiation emitted during the transition of electrons from state A to state B is :-

Solution & Explanation

Related Formula

The energy of the emitted photon during an atomic transition between energy states is given by:

Δ E = (hc)/(λ)

where h is Planck's constant, c is speed of light, and λ is the photon wavelength.

Core Logic

Write equations for the energy transitions from the layout of levels

Bohr Model of the Hydrogen Atom diagram for Q5 - JEE Main 2025 Morning
Bohr Model of the Hydrogen Atom diagram for Q5 - JEE Main 2025 Morning
:

EA - EC = hcλAC = hc2000 AA (i) EB - EC = hcλBC = hc6000 AA (ii)
Step 1: Finding Transition A to B

Subtracting equation (ii) from equation (i) gives the net transition energy from A to B :

EA - EB = (EA - EC) - (EB - EC) hcλAB = (hc)/(2000) - (hc)/(6000) 1λAB = (3 - 1)/(6000) = (2)/(6000) = (1)/(3000) λAB = 3000 AA
Pattern Recognition

Energy differences add linearly, which means their corresponding inverse wavelengths satisfy a parallel reciprocal subtraction rule: 1λAB = 1λAC - 1λBC.

Chapter Mix

Class 12 Physics: Atoms

Reference Study Guides

More Atoms Previous-Year Questions — Page 2

Q11 jee_main_2025_02_april_evening Bohr's Atomic Model
Assuming the validity of Bohr's atomic model for hydrogen like ions the radius of Li⁺⁺ ion in its ground state is given by 1Xa₀ , where X = _ _ _ . (Where a₀ is the first Bohr's radius.)
  • A. 2
  • B. 1
  • C. 3
  • D. 9

Solution

Related Formula

Bohr radius formula for hydrogen-like species:

rₙ = a₀ (n²)/(Z)
Core Logic

For Lithium ion Li⁺⁺:

  • Atomic number Z = 3
  • For ground state, the principal quantum number n = 1
  • Substitute Z = 3 and n = 1 into Bohr's radius formula:

r₁ = a₀ (1²)/(3) = (a₀)/(3)

Comparing with the given expression (1)/(X) a₀:

(1)/(X) a₀ = (a₀)/(3) X = 3
Pattern Recognition

Sees: Ground state radius of hydrogen-like species. Trap: Confusing the atomic number Z of Lithium with Helium (Z=2) or Beryllium (Z=4). Shortcut: Ground state radius of hydrogenic species is simply a₀ / Z. Since Lithium has Z=3, the ground state radius must be a₀/3 directly.

Chapter Mix

Class 12 Physics: Atoms

Q7 jee_main_2025_03_april_evening Bohr's Model of Hydrogen Atom
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The Bohr model is applicable to hydrogen and hydrogen-like atoms only. Reason R: The formulation of Bohr model does not include repulsive force between electrons. In the light of the above statements, choose the correct answer from the options given below :
  • A. Both A and R are true but R is NOT the correct explanation of A.
  • B. A is false but R is true.
  • C. Both A and R are true and R is the correct explanation of A.
  • D. A is true but R is false.

Solution

Related Formula

Bohr's electrostatic centripetal balance equation:

(m v²)/(r) = (1)/(4πε₀) (Z e²)/(r²)

This basic equation matches ONLY a single electron orbiting a nucleus of charge +Ze.

Core Logic

Assertion Analysis:

  • Bohr's model matches single-electron species (such as H, He^+, Li²⁺, etc.). Hence, Assertion A is true.
  • Reason Analysis:

  • The model is restricted because it models ONLY the attractive force between the positive nucleus and one orbiting electron. It cannot handle multi-electron systems due to the presence of inter-electron repulsive forces, which are not integrated into Bohr's simple formulation. Hence, Reason R is true.
  • Connection Check:

  • The lack of repulsive forces is exactly why the model fails for multi-electron species and remains applicable only to single-electron (hydrogen-like) species. Thus, R is the correct explanation of A.
Pattern Recognition

For single vs. multi-electron systems in atomic physics: Bohr model = single-electron ONLY. Quantum mechanics (Schrodinger) is required for multi-electron systems due to electron-electron interactions.

Chapter Mix

Class 12 Physics: Atoms

Q24 jee_main_2025_03_april_evening Hydrogen Spectrum and Energy Level Transitions
An electron in the hydrogen atom initially in the fourth excited state makes a transition to nth energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________.
Numerical Answer. Answer: 2 to 2

Solution

Related Formula

The energy of an electron in the n-th shell of a hydrogen atom is:

Eₙ = -(13.6)/(n²)~eV

The emitted photon energy during transition nᵢ arrow nf is:

Δ E = Enᵢ - Enf
Core Logic

Given state:

  • Initial state: "fourth excited state"
Step 1: Calculate energy of initial state ($E_5)
E₅ = -(13.6)/(5²) = -(13.6)/(25) = -0.544~eV
Step 2: Calculate final state energy ($Eₙ$)
2.86 = E_5 - E_n = -0.544 - E_nE_n = -0.544 - 2.86 = -3.404\mathrm{~eV}$
Step 3: Determine the integer shell index ($n$)
-3.4 = -\frac{13.6}{n^2}n^2 = \frac{13.6}{3.4} = 4 \Rightarrow n = 2$
Pattern Recognition

Memorizing the first few energy levels of the hydrogen atom saves calculation time:

Q15 jee_main_2025_07_april_morning Bohr Model of Hydrogen Atom
In a hydrogen like ion, the energy difference between the 2nd excitation energy state and ground is 108.8eV . The atomic number of the ion is
  • A. 4
  • B. 2
  • C. 1
  • D. 3

Solution

Related Formula

For a hydrogen-like ion of atomic number Z, the transition energy between states n₂ and n₁ is:

Δ E = 13.6 Z² ( (1)/(n₁²) - (1)/(n₂²) ) ~eV
Core Logic

Identify the states:

  • Ground state:
  • Substitute values into the energy equation:

$
108.8 = 13.6 Z² ( (1)/(1²) - (1)/(3²) )108.8 = 13.6 Z² ( 1 - (1)/(9) ) = 13.6 Z² × (8)/(9)
Step 1: Solve for Z

Rearrange the equation to isolate

Step 1: Solve for Z

Rearrange the equation to isolate $Z^2:

Z² = (108.8 × 9)/(13.6 × 8)

Calculate the numerical value:

(108.8)/(13.6) = 8Z² = (8 × 9)/(8) = 9 Z = 3
Pattern Recognition

Sees: "

Pattern Recognition

Sees: "$2^{\mathrm{nd}}excitation"\implies n = 3. Ground state\implies n = 1. Shortcut: Energy transition ratio for1 \to 3is always\frac{8}{9}of13.6 Z^2 \approx 12.1 Z^2. Since108.8 / 12.1 \approx 9,Z^2 = 9 \implies Z = 3$.

Chapter Mix

Class 12 Physics: Atoms

Q16 jee_main_2025_07_april_morning Hydrogen Spectrum
For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.
  • A. 5:36
  • B. 5:27
  • C. 3 : 4
  • D. 27:5

Solution

Related Formula

The wavelength λ for a transition in a hydrogen atom is given by the Rydberg formula:

(1)/(λ) = R ( (1)/(n₁²) - (1)/(n₂²) )

To find the largest wavelength (minimum energy transition), select the adjacent higher shell n₂ = n₁ + 1.

Core Logic
  • Lyman Series largest wavelength (λL): Transition from n = 2 → 1
(1)/(λL) = R ( (1)/(1²) - (1)/(2²) ) = (3R)/(4) λL = (4)/(3R)
  • Balmer Series largest wavelength (λB): Transition from n = 3 → 2
(1)/(λB) = R ( (1)/(2²) - (1)/(3²) ) = R ( (1)/(4) - (1)/(9) ) = (5R)/(36) λB = (36)/(5R)
Step 1: Ratio Calculation

Now, compute the ratio of the wavelengths:

(λL)/(λB) = ((4)/(3R))/((36)/(5R)) = (4)/(3) × (5)/(36) = (5)/(27)
Pattern Recognition

Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level

Pattern Recognition

Sees: Ratio of largest wavelengths of series. Shortcut: The largest wavelength in a series starting at ground level $n_1is\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}. For Lyman (n_1=1):\lambda_L \propto \frac{4}{3}. For Balmer (n_1=2):\lambda_B \propto \frac{36}{5}. Ratio:\frac{4/3}{36/5} = \frac{5}{27}$.

Chapter Mix

Class 12 Physics: Atoms

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