For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Year	2026	2025	2024	Total
Questions	5	8	9	22

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

The wavelength $λ$ $\lambda$ 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 $n₂ = n₁ + 1$ $n_2 = n_1 + 1$.

Core Logic

Lyman Series largest wavelength ( $λL$ $\lambda_L$): Transition from $n = 2 \to 1$ $n = 2 \to 1$

(1)/(λL) = R ( (1)/(1²) - (1)/(2²) ) = (3R)/(4) λL = (4)/(3R)

Balmer Series largest wavelength ( $λB$ $\lambda_B$): Transition from $n = 3 \to 2$ $n = 3 \to 2$

(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) \times (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 $n_1is$ is $\lambda \propto \frac{n_1^2 (n_1+1)^2}{2n_1 + 1}. For Lyman ($. For Lyman ($n_1=1):$): $\lambda_L \propto \frac{4}{3}. For Balmer ($. For Balmer ($n_1=2):$): $\lambda_B \propto \frac{36}{5}. Ratio:$. Ratio: $\frac{4/3}{36/5} = \frac{5}{27}$. Chapter Mix Class 12 Physics: Atoms

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 $n$ $n$ corresponding to the energy $-0.85 ~eV$ $-0.85 \mathrm{~eV}$. 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

$n = 4$ $n = 4$

Step 2: Calculate Transitions

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

= (n(n - 1))/(2)

= (4(4 - 1))/(2) = (12)/(2) = 6

Pattern Recognition

Energy states in Hydrogen are heavily standardized: $n=1 arrow -13.6$ $n=1 \rightarrow -13.6$, $n=2 arrow -3.4$ $n=2 \rightarrow -3.4$, $n=3 arrow -1.51$ $n=3 \rightarrow -1.51$, $n=4 arrow -0.85$ $n=4 \rightarrow -0.85$. Recognize $-0.85 ~eV$ $-0.85 \mathrm{~eV}$ 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. $(27)/(32) λ$ $\frac{27}{32} \lambda$
B. $(32)/(27)λ$ $\frac{32}{27}\lambda$
C. $(27)/(5) λ$ $\frac{27}{5} \lambda$
D. $(5)/(27)λ$ $\frac{5}{27}\lambda$

Solution

Related Formula

(1)/(λ) = R Z² [ (1)/(n₁²) - (1)/(n₂²) ]

Core Logic

For the first member of the Lyman series of hydrogen ( $n₁ = 1, n₂ = 2$ $n_1 = 1, n_2 = 2$):

(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 ( $n₁ = 1, n₂ = 3$ $n_1 = 1, n_2 = 3$):

(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) \times (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 $[1/n₁² - 1/n₂²]$ $[1/n_1^2 - 1/n_2^2]$. For Lyman 1st and 2nd, the ratio is $(3/4) / (8/9) = 27/32$ $(3/4) / (8/9) = 27/32$.

Chapter Mix

Class 12 Physics: Atoms

For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Solution & Explanation

Related Formula

Core Logic

Step 1: Ratio Calculation

Pattern Recognition

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Find Quantum State

Step 2: Calculate Transitions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Second Member Calculation

Step 3: Ratio

Pattern Recognition

Chapter Mix

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For a hydrogen atom, the ratio of the largest wavelength of Lyman series to that of the Balmer series is.

Solution & Explanation

Related Formula

Core Logic

Step 1: Ratio Calculation

Pattern Recognition

Pattern Recognition

Chapter Mix

Reference Study Guides

More Atoms Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Find Quantum State

Step 2: Calculate Transitions

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 2: Second Member Calculation

Step 3: Ratio

Pattern Recognition

Chapter Mix

More Atoms Questions — jee_main_2025_07_april_morning

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