Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer…

Q: Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series The correct

Let's find the values of n1 and n2 for each transition: - (A) Paschen (1st line): n1 = 3, n2 = 4 - (B) Balmer (2nd line): n1 = 2, n2 = 4 - (C) Paschen (3rd line): n1 = 3, n2 = 6 - (D) Bracket (4th line): n1 = 4, n2 = 8 Calculating or comparing the energy values corresponding to these transitions yields the ascending order of energy. The correct ascending order of energy of the given lines is D < A < C < B, corresponding to option (4). Sees: hydrogen spectral lines energy comparison. Trap: Confus

Year	2026	2025	2024	Total
Questions	14	16	9	39

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

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

Core Logic

Let's find the values of $n₁$ $n_1$ and $n₂$ $n_2$ for each transition:

(A) Paschen ( $1st$ $1^{st}$ line): $n₁ = 3, n₂ = 4$ $n_1 = 3, n_2 = 4$
(B) Balmer ( $2nd$ $2^{nd}$ line): $n₁ = 2, n₂ = 4$ $n_1 = 2, n_2 = 4$
(C) Paschen ( $3rd$ $3^{rd}$ line): $n₁ = 3, n₂ = 6$ $n_1 = 3, n_2 = 6$
(D) Bracket ( $4th$ $4^{th}$ line): $n₁ = 4, n₂ = 8$ $n_1 = 4, n_2 = 8$

Calculating or comparing the energy values corresponding to these transitions yields the ascending order of energy.

Step 1: Final Conclusion

The correct ascending order of energy of the given lines is $D < A < C < B$ $D < A < C < B$, corresponding to option (4).

Pattern Recognition

Sees: hydrogen spectral lines energy comparison. Trap: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents.

Chapter Mix

Class 11 Chemistry: 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 $Δ n = n₂ - n₁$ $\Delta n = n_2 - n_1$

Core Logic

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

Fifth excited state means $n₂ = 5 + 1 = 6$ $n_2 = 5 + 1 = 6$.
First excited state means $n₁ = 1 + 1 = 2$ $n_1 = 1 + 1 = 2$.

Calculate the difference:

Δ n = n₂ - n₁ = 6 - 2 = 4

Step 1: Calculate Total Spectral Lines

Substitute $Δ n = 4$ $\Delta n = 4$ into the formula:

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

= (4 \times 5)/(2)

$= 10$ $= 10$

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 ( $1s¹$ $1s^1$), the energy of an orbital depends only on the principal quantum number ' $n$ $n$'.

Step 1: Final conclusion

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

Pattern Recognition

Hydrogen atom (single electron species) = Energy dictated strictly by $n$ $n$. Multi-electron atoms = Energy dictated by $(n + l)$ $(n + l)$ 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 $19$ $19$. Its electronic configuration is: $1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹$ $1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost orbital of potassium is the $4s$ $4s$ orbital.

For the $4s$ $4s$ orbital: Principal quantum number, $n = 4$ $n = 4$ Azimuthal quantum number, $l = 0$ $l = 0$ (for s-subshell) Magnetic quantum number, $m = 0$ $m = 0$ (since $l = 0$ $l = 0$) Spin quantum number, $s = \pm (1)/(2)$ $s = \pm \frac{1}{2}$

Step 1: Final Matching

Comparing these values with the options, option (2) provides $n = 4, l = 0, m = 0, s = +(1)/(2)$ $n = 4, l = 0, m = 0, s = +\frac{1}{2}$, 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 \times NA

Step 1: Energy per atom

E = (1240)/(242) eV = 5.12 eV

Converting eV to Joules:

E = 5.12 \times 1.6 \times 10⁻¹⁹ J/atom

E = 8.192 \times 10⁻¹⁹ J/atom

Step 2: Energy per mole

To find the ionization energy per mole, multiply by Avogadro's number ( $NA \approx 6.022 \times 10²³$ $N_A \approx 6.022 \times 10^{23}$):

IE = 8.192 \times 10⁻¹⁹ \times 6.022 \times 10²³ J/mol

IE \approx 493.3 \times 10³ J/mol \approx 494 kJ/mol

Chapter Mix

Class 11 Chemistry: Structure of Atom

Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series The correct arrangement of the above lines in ascending order of energy is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Total Spectral Lines

Chapter Mix

Solution

Core Logic

Step 1: Final conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Matching

Chapter Mix

Solution

Related Formula

Step 1: Energy per atom

Step 2: Energy per mole

Chapter Mix

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Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series The correct arrangement of the above lines in ascending order of energy is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Calculate Total Spectral Lines

Chapter Mix

Solution

Core Logic

Step 1: Final conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Matching

Chapter Mix

Solution

Related Formula

Step 1: Energy per atom

Step 2: Energy per mole

Chapter Mix

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