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:

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: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents. 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

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

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:

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: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents. 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

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

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

Year	2026	2025	2024	Total
Questions	14	16	9	39

Q61 jee_main_2026_23_january_morning Quantum Numbers

Given, (A)

n = 5, ml = -1

(B)

n = 3, l = 2, ml = -1, mₛ = +(1)/(2)

The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are:

A. $26 and 1$ $\text{26 and 1}$
B. $4 and 1$ $\text{4 and 1}$
C. $2 and 4$ $\text{2 and 4}$
D. $8 and 1$ $\text{8 and 1}$

Solution

Core Logic

Evaluate constraints set by the given quantum numbers to find available electron slots. Each unique orbital $(n, l, ml)$ $(n, l, m_l)$ can hold exactly 2 electrons of opposite spin.

Step 1: Constraint A

For (A) $n = 5$ $n = 5$, $ml = -1$ $m_l = -1$: The possible azimuthal quantum numbers $l$ $l$ range from $0$ $0$ to $n-1 = 4$ $n-1 = 4$. However, $ml$ $m_l$ goes from $-l$ $-l$ to $+l$ $+l$. For $ml = -1$ $m_l = -1$ to exist, $l$ $l$ must be at least $1$ $1$. Possible $l$ $l$ values: $l=1, l=2, l=3, l=4$ $l=1, l=2, l=3, l=4$. Each of these subshells ( $5p, 5d, 5f, 5g$ $5p, 5d, 5f, 5g$) contains exactly one orbital where $ml = -1$ $m_l = -1$. Total orbitals = $4$ $4$. Total electrons = $4 orbitals \times 2 e^- /orbital = 8 electrons$ .

Step 2: Constraint B

For (B) $n = 3, l = 2, ml = -1, mₛ = +(1)/(2)$ : This completely specifies all four quantum numbers for a single state. Pauli's Exclusion Principle states no two electrons can have the same four quantum numbers. Therefore, exactly 1 electron is possible.

Pattern Recognition

If $n$ $n$ and $ml \neq 0$ $m_l \neq 0$ are given, count how many $l$ $l$ values are $\geq |ml|$ $\ge |m_l|$. For $n=5$ $n=5$ and $ml=-1$ $m_l=-1$, $l in 1,2,3,4$ $l \in \{1,2,3,4\}$. That's 4 orbitals, 8 electrons.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q64 jee_main_2026_23_january_evening Planck's Quantum Theory and Photoelectric Effect

Identify the INCORRECT statements from the following: A. Notation

²⁴₁₂Mg

represents 24 protons and 12 neutrons. B. Wavelength of a radiation of frequency

4.5 \times 10¹⁵ s⁻¹

is

6.7 \times 10⁻⁸ m

. C. One radiation has wavelength =

λ₁(900 nm)

and energy =

E₁

. Other radiation has wavelength =

λ₂(300 nm)

and energy =

E₂

.

E₁ : E₂ = 3 : 1

. D. Number of photons of light of wavelength

2000 pm

that provides

1 J

of energy is

1.006 \times 10¹⁶

. Choose the correct answer from the options given below:

A. $A and D only$ $\text{A and D only}$
B. $A and C only$ $\text{A and C only}$
C. $A and B only$ $\text{A and B only}$
D. $B and C only$ $\text{B and C only}$

Solution

Related Formula

$c = λ ν$ $c = \lambda \nu$

E = (hc)/(λ)

Etotal = n (hc)/(λ)

Core Logic

Evaluate each statement:

A. The notation $²⁴₁₂Mg$ ${}^{24}_{12}Mg$ specifies atomic number $Z = 12$ $Z = 12$ (12 protons) and mass number $A = 24$ $A = 24$ (protons + neutrons). Number of neutrons = $24 - 12 = 12$ $24 - 12 = 12$. The statement incorrectly says 24 protons. (False)

B. Wavelength $λ = (c)/(ν) = 3 \times 10⁸4.5 \times 10¹⁵ = 0.666 \times 10⁻⁷ m = 6.67 \times 10⁻⁸ m$ . Statement B is approximately correct. (True)

C. Energy is inversely proportional to wavelength. $E₁ \propto (1)/(λ₁)$ $E_1 \propto \frac{1}{\lambda_1}$ and $E₂ \propto (1)/(λ₂)$ $E_2 \propto \frac{1}{\lambda_2}$. $(E₁)/(E₂) = (λ₂)/(λ₁) = (300)/(900) = (1)/(3)$ . The statement claims $E₁ : E₂ = 3:1$ $E_1 : E_2 = 3:1$, which is incorrect. (False)

D. Number of photons $n = Etotal \times λhc = 1 \times 2000 \times 10⁻¹²6.626 \times 10⁻³⁴ \times 3 \times 10⁸ \approx 1.006 \times 10¹⁶$ . (True)

Step 1: Conclusion

The incorrect statements are A and C.

Pattern Recognition

Longer wavelength means lower energy. A wavelength 3x larger means energy is exactly 1/3x. Don't flip the inverse proportionality ratio.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q68 jee_main_2026_23_january_evening Photoelectric Effect

The work functions of two metals (

MA

and

MB

) are in the

1:2

ratio. When these metals are exposed to photons of energy

6 eV

, the kinetic energy of liberated electrons of

MA:MB

is in the ratio of

2.642:1

. The work functions (in

eV

) of

MA

and

MB

are respectively.

A. $3.1, 6.2$ $3.1, 6.2$
B. $2.3, 4.6$ $2.3, 4.6$
C. $1.4, 2.8$ $1.4, 2.8$
D. $1.5, 3.0$ $1.5, 3.0$

Solution

Related Formula

KEmax = E - φ

where $E$ $E$ is photon energy and $φ$ $\phi$ is the work function.

Core Logic

Given the ratio of work functions is $1:2$ $1:2$, let the work function of $MA$ $M_A$ be $φ₁$ $\phi_1$ and $MB$ $M_B$ be $φ₂$ $\phi_2$. Then $φ₂ = 2φ₁$ $\phi_2 = 2\phi_1$.

For metal $MA$ $M_A$ exposed to $6 eV$ $6 \text{ eV}$ photons: $(KEmax)₁ = 6 - φ₁$ $(KE_{\text{max}})_1 = 6 - \phi_1$ --- (Equation 1)

For metal $MB$ $M_B$ exposed to $6 eV$ $6 \text{ eV}$ photons: $(KEmax)₂ = 6 - φ₂ = 6 - 2φ₁$ --- (Equation 2)

Step 1: Setting up the ratio

We are given the kinetic energy ratio:

(KEmax)₁(KEmax)₂ = (2.642)/(1)

Substitute equations 1 and 2 into the ratio:

(6 - φ₁)/(6 - 2φ₁) = 2.642

Step 2: Solving for work function

6 - φ₁ = 2.642 \times (6 - 2φ₁)

6 - φ₁ = 15.852 - 5.284φ₁

5.284φ₁ - φ₁ = 15.852 - 6

4.284φ₁ = 9.852

φ₁ = (9.852)/(4.284) = 2.3 eV

Since $φ₂ = 2φ₁$ $\phi_2 = 2\phi_1$:

φ₂ = 2 \times 2.3 = 4.6 eV

Pattern Recognition

Photoelectric effect ratio problems are purely algebraic setups. Express all unknowns in terms of a single variable using the given multiplier ( $1:2$ $1:2$ ratio) and plug directly into Einstein's photoelectric equation.

Chapter Mix

Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter

Q73 jee_main_2026_24_january_morning Hydrogen Spectrum

The hydrogen spectrum consists of several spectral lines in Lyman series (

L₁

,

L₂

,

L₃

....;

L₁

has lowest energy among Lyman series). Similarly it consists of several spectral lines in Balmer series (

B₁

,

B₂

,

B₃

..... ;

B₁

has lowest energy among Balmer lines). The energy of

L₁

is x times the energy of

B₁

. The value of x is ____

\times 10⁻¹

(Nearest integer)

Numerical Answer. Answer: 54 to 54

Solution

Related Formula

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

Core Logic

Lowest energy line in Lyman series ( $L₁$ $L_1$) corresponds to transition from $n₂ = 2$ $n_2 = 2$ to $n₁ = 1$ $n_1 = 1$. $Δ E(L₁) = 13.6 \times Z² ( (1)/(1²) - (1)/(2²) ) = 13.6 \times Z² \times (3)/(4)$

Lowest energy line in Balmer series ( $B₁$ $B_1$) corresponds to transition from $n₂ = 3$ $n_2 = 3$ to $n₁ = 2$ $n_1 = 2$. $Δ E(B₁) = 13.6 \times Z² ( (1)/(2²) - (1)/(3²) ) = 13.6 \times Z² \times ( (1)/(4) - (1)/(9) ) = 13.6 \times Z² \times (5)/(36)$

Step 1: Find Ratio x

(Δ E(L₁))/(Δ E(B₁)) = (13.6 Z² \times 3/4)/(13.6 Z² \times 5/36)

x = (3/4)/(5/36) = (3)/(4) \times (36)/(5) = (3 \times 9)/(5) = (27)/(5) = 5.4

We need to express $x$ $x$ as $____ \times 10⁻¹$ $____ \times 10^{-1}$. $5.4 = 54 \times 10⁻¹$ $5.4 = 54 \times 10^{-1}$ Thus, the integer value is $54$ $54$.

Pattern Recognition

Lowest energy always means the transition from the immediately adjacent higher level ( $n+1 arrow n$ $n+1 \rightarrow n$). Set up the ratio to immediately cancel out the Rydberg constant/13.6 term.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q63 jee_main_2026_24_january_evening Bohr Model for Hydrogen-like Species

The wavelength of spectral line obtained in the spectrum of

LI²⁺

ion, when the transition takes place between two levels whose sum is 4 and difference is 2, is

A. $2.28 \times 10⁻⁷ ~cm$ $2.28 \times 10^{-7} \mathrm{~cm}$
B. $2.28 \times 10⁻⁶ ~cm$ $2.28 \times 10^{-6} \mathrm{~cm}$
C. $1.14 \times 10⁻⁷ ~cm$ $1.14 \times 10^{-7} \mathrm{~cm}$
D. $1.14 \times 10⁻⁶ ~cm$ $1.14 \times 10^{-6} \mathrm{~cm}$

Solution

Related Formula

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

Core Logic

Let $n₁$ $n_1$ be the lower energy level and $n₂$ $n_2$ be the higher energy level. Given conditions: $n₁ + n₂ = 4$ $n_1 + n_2 = 4$ $n₂ - n₁ = 2$ $n_2 - n_1 = 2$

Solving these linear equations: $2n₂ = 6 n₂ = 3$ $2n_2 = 6 \implies n_2 = 3$ $n₁ = 1$ $n_1 = 1$

Step 1: Apply Rydberg Formula

For $Li²⁺$ $Li^{2+}$, the atomic number $Z = 3$ $Z = 3$.

(1)/(λ) = RH (3)² [ (1)/(1²) - (1)/(3²) ]

(1)/(λ) = 9 RH [ 1 - (1)/(9) ]

(1)/(λ) = 9 RH \times (8)/(9) = 8RH

λ = 18RH

Step 2: Calculate Wavelength in cm

Using $RH = 109677 ~cm⁻¹ 1.1 \times 10⁵ ~cm⁻¹$ (approximated for calculation):

λ = (1)/(8 \times 1.1 \times 10⁵)

λ = (1)/(8.8 \times 10⁵)

λ = (1000)/(8.8) \times 10⁻⁸ ~cm

λ = 113.63 \times 10⁻⁸ ~cm

λ 1.14 \times 10⁻⁶ ~cm

Pattern Recognition

Equations translating "sum and difference of levels" instantly lock in the specific transition ( $3 arrow 1$ $3 \rightarrow 1$ Lyman series line). Always check the $Z²$ $Z^2$ factor for hydrogen-like species (like $Li²⁺$ $Li^{2+}$ where $Z²=9$ $Z^2=9$) which heavily alters the wavelength.

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 2

Solution

Core Logic

Step 1: Constraint A

Step 2: Constraint B

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting up the ratio

Step 2: Solving for work function

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Ratio x

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Rydberg Formula

Step 2: Calculate Wavelength in cm

Pattern Recognition

Chapter Mix

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