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Atoms appeared 22 times across 3 years — 2.5% of Physics. This question is from Hydrogen Spectrum.

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The minimum energy required by a hydrogen atom in ground state to emit radiation in Balmer series is nearly:

Solution & Explanation

Related Formula

Bohr state energy level values:

Eₙ = -(13.6)/(n²)~eV

Transition excitation requirement:

Δ E = Efinal - Einitial
Core Logic

To emit radiation in the Balmer series, the hydrogen electron must first be excited to at least the n=3 shell. This allows it to jump down to n=2 and produce the first spectral line of the Balmer series.

Energy required to transition from ground state (n=1) to n=3:

E₁ = -13.6~eV E₃ = -(13.6)/(3²) = -1.51~eV
Step 1: Calculate Energy Gap
Δ E = E₃ - E₁ = -1.51 - (-13.6) = 12.09~eV ≈ 12.1~eV
Pattern Recognition

Balmer emissions always return down to n=2. Thus, the initial excitation starting from ground state n=1 must reach at least n=3 to create a valid Balmer transition.

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Class 12 Physics: Atoms

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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.

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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.

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Class 12 Physics: Atoms

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