The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is 1016 s-1. The wave number of photon 'C' is 104…

Q: The wavelength of photon 'A' is 400 nm. The frequency of photon 'B' is 1016 s-1. The wave number of photon 'C' is 104 cm-1. The correct order of energy of these photons is :

Let's convert all parameters to wavelength (lambda) in nanometers for easy comparison. (1) Wavelength of A = 400 nm. (2) Wavelength of B (lambdaB): (3) Wavelength of C (lambdaC): Comparing wavelengths: lambdaC (1000 nm) > lambdaA (400 nm) > lambdaB (30 nm) Since Energy (E) is inversely proportional to wavelength (E propto (1/lambda)): EB > EA > EC Convert distinct units (frequency, wavenumber, wavelength) into a single metric (usually wavelength) to rank. Remember E propto 1/lambda. Class 11 Che

Year	2026	2025	2024	Total
Questions	14	16	9	39

Q jee_main_2024_01_february_morning Dual Behaviour of Matter

According to the wave-particle duality of matter by de-Broglie, which of the following graph plot presents most appropriate relationship between wavelength of electron (

λ

) and momentum of electron (

p

)?

A.
B.
C.
D.

Solution

Related Formula

λ = (h)/(p)

Core Logic

From the de-Broglie equation: $λ \propto (1)/(p)$ $\lambda \propto \frac{1}{p}$ $\Rightarrow λ p = h (constant)$

This represents the equation of a rectangular hyperbola ( $xy = c$ $xy = c$).

Step 1: Graph Identification

The plot of $λ$ $\lambda$ versus $p$ $p$ will be a rectangular hyperbola curve in the first quadrant. Graph 1 correctly depicts this hyperbolic relationship.

Dual Behaviour of Matter diagram for Q66 - JEE Main 2024 Morning

Pattern Recognition

Inverse proportionality $y = (k)/(x)$ $y = \frac{k}{x}$ always graphs as a rectangular hyperbola in the positive quadrant.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q jee_main_2024_29_january_evening Hydrogen Spectral Series

Match List I with List II:

List I (Spectral Series for Hydrogen)	List II (Spectral Region/Higher Energy State)
A. Lyman	I. Infrared region
B. Balmer	II. UV region
C. Paschen	III. Infrared region
D. Pfund	IV. Visible region

Choose the correct answer from the options given below :

A. A-II, B-III, C-I, D-IV
B. A-I, B-III, C-II, D-IV
C. A-II, B-IV, C-III, D-I
D. A-I, B-II, C-III, D-IV

Solution

Related Formula

(1)/(λ) = RH ( (1)/(n₁²) - (1)/(n₂²) )

Core Logic

Analyzing electromagnetic spectrum regions for each hydrogen electronic decay series:

Lyman Series ( $n₁=1$ $n_1=1$) emits photons in the high-energy Ultraviolet (UV) region (II).
Balmer Series ( $n₁=2$ $n_1=2$) falls cleanly into the Visible region spectrum (IV).
Paschen Series ( $n₁=3$ $n_1=3$) transitions sit inside the standard Near-Infrared region (III).
Pfund Series ( $n₁=5$ $n_1=5$) shifts sit even deeper in the Infrared region zone (I).

Step 1: Resolution

Thus, compiling match mappings yields configuration: A-II, B-IV, C-III, D-I.

Pattern Recognition

Lyman is always UV; Balmer is visible; all subsequent higher level series ( $n₁ \geq 3$ $n_1 \ge 3$) belong strictly to infrared segments.

Chapter Mix

Class 11 Chemistry: Atomic Structure

Q jee_main_2024_27_jan_morning Quantum Numbers and Electron Capacity

The number of electrons present in all the completely filled subshells having

n=4

and

s=+(1)/(2)

is

. (Where

n=

principal quantum number and

s=

spin quantum number)

Numerical Answer. Answer: 16 to 16

Solution

Step 1: Identify all available subshells within the $n=4$ $n=4$ energy shell

For principal quantum level $n=4$ $n=4$, the allowed values of azimuthal quantum numbers ( $l$ $l$) are:

$4s$ $4\text{s}$ ( $l=0$ $l=0$) $arrow 1 orbital arrow 2 electrons capacity$
$4p$ $4\text{p}$ ( $l=1$ $l=1$) $arrow 3 orbitals arrow 6 electrons capacity$
$4d$ $4\text{d}$ ( $l=2$ $l=2$) $arrow 5 orbitals arrow 10 electrons capacity$
$4f$ $4\text{f}$ ( $l=3$ $l=3$) $arrow 7 orbitals arrow 14 electrons capacity$

Step 2: Filter capacity using spin values

Every single spatial orbital holds exactly 2 electrons maximum; one with spin $s=+(1)/(2)$ $s=+\frac{1}{2}$ and one with spin $s=-(1)/(2)$ $s=-\frac{1}{2}$. Total number of orbitals across $n=4$ $n=4$ is:

1 + 3 + 5 + 7 = 16 orbitals

Thus, the total count of electrons featuring spin value $s=+(1)/(2)$ $s=+\frac{1}{2}$ across these completely filled configurations is exactly $16$ $16$.

Pattern Recognition

Total orbitals in shell $n$ $n$ is $n²$ $n^2$. Since each orbital contributes exactly 1 electron with $s=+(1)/(2)$ $s=+\frac{1}{2}$, capacity is simply $n² = 4² = 16$ $n^2 = 4^2 = 16$.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q65 jee_main_2024_27_jan_morning Electronic Configuration and Magnetic Moment

Which of the following electronic configuration would be associated with the highest magnetic moment?

A. $[Ar] 3d⁷$ $[Ar] 3d^7$
B. $[Ar] 3d⁸$ $[Ar] 3d^8$
C. $[Ar] 3d³$ $[Ar] 3d^3$
D. $[Ar] 3d⁶$ $[Ar] 3d^6$

Solution

Related Formula

Spin-only magnetic moment formula:

μ = \sqrt(n(n+2)) BM

where $n$ $n$ is the number of unpaired electrons.

Step 1: Audit configurations and count unpaired electrons

Configuration	Unpaired $e^-$ $e^-$ ( $n$ $n$)	Magnetic Moment (BM)
$[Ar] 3d⁷$ $[Ar] 3d^7$	3	$\sqrt(15)$ $\sqrt{15}$
$[Ar] 3d⁸$ $[Ar] 3d^8$	2	$\sqrt(8)$ $\sqrt{8}$
$[Ar] 3d³$ $[Ar] 3d^3$	3	$\sqrt(15)$ $\sqrt{15}$
$[Ar] 3d⁶$ $[Ar] 3d^6$	4	$\sqrt(24)$ $\sqrt{24}$

Step 2: Conclusion

Since $[Ar] 3d⁶$ $[Ar] 3d^6$ contains the maximal count of 4 unpaired electrons, it yields the highest spin-only magnetic moment value.

Pattern Recognition

Maximal unpaired configuration in high-spin $3dⁿ$ $3d^n$ series yields the highest $μ$ $\mu$. Check $n$ $n$ systematically.

Chapter Mix

Class 11 Chemistry: Structure of Atom Class 12 Chemistry: d-and f-Block Elements

Q66 jee_main_2024_29_jan_morning Quantum Numbers

The correct set of four quantum numbers for the valence electron of rubidium atom

(Z = 37)

is:

A. $5, 0, 0, +(1)/(2)$ $5, 0, 0, +\frac{1}{2}$
B. $5, 0, 1, +(1)/(2)$ $5, 0, 1, +\frac{1}{2}$
C. $5, 1, 0, +(1)/(2)$ $5, 1, 0, +\frac{1}{2}$
D. $5, 1, 1, +(1)/(2)$ $5, 1, 1, +\frac{1}{2}$

Solution

Core Logic

Rubidium ( $Rb$ $Rb$) has the atomic number $Z = 37$ $Z = 37$.

The noble gas core preceding it is Krypton ( $Kr$ $Kr$, $Z = 36$ $Z = 36$). The electronic configuration is:

Rb arrow [Kr] 5s¹

Thus, the valence electron resides in the $5s$ $5s$ orbital.

Step 1: Assigning Quantum Numbers

For a $5s$ $5s$ electron:

Principal quantum number, $n = 5$ $n = 5$ (from the shell number).
Azimuthal quantum number, $l = 0$ $l = 0$ (for an s-orbital).
Magnetic quantum number, $m = 0$ $m = 0$ (since $m$ $m$ ranges from $-l$ $-l$ to $+l$ $+l$, and $l=0$ $l=0$).
Spin quantum number, $s = +(1)/(2)$ $s = +\frac{1}{2}$ or $-(1)/(2)$ $-\frac{1}{2}$.

The correct set matching this is $5, 0, 0, +(1)/(2)$ $5, 0, 0, +\frac{1}{2}$.

Chapter Mix

Class 11 Chemistry: Structure of Atom

The wavelength of photon 'A' is 400 nm$400\text{ nm}$. The frequency of photon 'B' is 10¹⁶ s⁻¹$10^{16}\text{ s}^{-1}$. The wave number of photon 'C' is 10⁴ cm⁻¹$10^{4}\text{ cm}^{-1}$. The correct order of energy of these photons is :

Solution & Explanation

Related Formula

Core Logic

Step 1: Compare Wavelengths and Energies

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Graph Identification

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Resolution

Pattern Recognition

Chapter Mix

Solution

Step 1: Identify all available subshells within the n=4$n=4$ energy shell

Step 2: Filter capacity using spin values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Audit configurations and count unpaired electrons

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Assigning Quantum Numbers

Chapter Mix

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The wavelength of photon 'A' is $400 nm$ $400\text{ nm}$. The frequency of photon 'B' is $10¹⁶ s⁻¹$ $10^{16}\text{ s}^{-1}$. The wave number of photon 'C' is $10⁴ cm⁻¹$ $10^{4}\text{ cm}^{-1}$. The correct order of energy of these photons is :

Step 1: Identify all available subshells within the $n=4$ $n=4$ energy shell