The energy of an electron in the first Bohr orbit of the Hydrogen atom is -13.6 eV. The magnitude of the energy value…

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Questions	14	15	9	38

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

The energy of an electron in the first Bohr orbit of the Hydrogen atom is -13.6 eV$-13.6 \text{ eV}$. The magnitude of the energy value of an electron in the first excited state of the Be³⁺$\text{Be}^{3+}$ ion is _________ eV (as the nearest integer value).

Solution & Explanation

Related Formula

Execution

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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The energy of an electron in the first Bohr orbit of the Hydrogen atom is $-13.6 eV$ $-13.6 \text{ eV}$. The magnitude of the energy value of an electron in the first excited state of the $Be³⁺$ $\text{Be}^{3+}$ ion is _________ eV (as the nearest integer value).