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:

Evaluate constraints set by the given quantum numbers to find available electron slots. Each unique orbital (n, l, ml) can hold exactly 2 electrons of opposite spin. For (A) n = 5, ml = -1: The possible azimuthal quantum numbers l range from 0 to n-1 = 4. However, ml goes from -l to +l. For ml = -1 to exist, l must be at least 1. Possible l values: l=1, l=2, l=3, l=4. Each of these subshells (5p, 5d, 5f, 5g) contains exactly one orbital where ml = -1. Total orbitals = 4. Total electrons = 4 orbitals x 2 e^- /orbital = 8 electrons. 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. If n and ml neq 0 are given, count how many l values are |ml|. For n=5 and ml=-1, l 1,2,3,4. That's 4 orbitals, 8 electrons. Class 11 Chemistry: Structure of Atom

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:

Evaluate constraints set by the given quantum numbers to find available electron slots. Each unique orbital (n, l, ml) can hold exactly 2 electrons of opposite spin. For (A) n = 5, ml = -1: The possible azimuthal quantum numbers l range from 0 to n-1 = 4. However, ml goes from -l to +l. For ml = -1 to exist, l must be at least 1. Possible l values: l=1, l=2, l=3, l=4. Each of these subshells (5p, 5d, 5f, 5g) contains exactly one orbital where ml = -1. Total orbitals = 4. Total electrons = 4 orbi

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:

Evaluate constraints set by the given quantum numbers to find available electron slots. Each unique orbital (n, l, ml) can hold exactly 2 electrons of opposite spin. For (A) n = 5, ml = -1: The possible azimuthal quantum numbers l range from 0 to n-1 = 4. However, ml goes from -l to +l. For ml = -1 to exist, l must be at least 1. Possible l values: l=1, l=2, l=3, l=4. Each of these subshells (5p, 5d, 5f, 5g) contains exactly one orbital where ml = -1. Total orbitals = 4. Total electrons = 4 orbitals x 2 e^- /orbital = 8 electrons. 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. If n and ml neq 0 are given, count how many l values are |ml|. For n=5 and ml=-1, l 1,2,3,4. That's 4 orbitals, 8 electrons. Class 11 Chemistry: Structure of Atom

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:

Evaluate constraints set by the given quantum numbers to find available electron slots. Each unique orbital (n, l, ml) can hold exactly 2 electrons of opposite spin. For (A) n = 5, ml = -1: The possible azimuthal quantum numbers l range from 0 to n-1 = 4. However, ml goes from -l to +l. For ml = -1 to exist, l must be at least 1. Possible l values: l=1, l=2, l=3, l=4. Each of these subshells (5p, 5d, 5f, 5g) contains exactly one orbital where ml = -1. Total orbitals = 4. Total electrons = 4 orbi

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…

Q70 jee_main_2026_21_jan_morning Hydrogen Spectrum

Given below are two statements: Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies. Statement II: The frequency of second line of Balmer series obtained from

He^+

is equal to that of first line of Lyman series obtained from hydrogen atom. In the light of the above statements, choose the correct answer from the options given below:

A. $textBoth Statement I and Statement II are true$
B. $textBoth Statement I and Statement II are false$
C. $textStatement I is false but Statement II is true$
D. $textStatement I is true but Statement II is false$

Solution

### Related Formula

frac1lambda = RZ^2left(frac1n_1^2 - frac1n_2^2right)

### Core Logic Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true. For Statement II: First line of Lyman series for H atom (

Z=1, n_1=1, n_2=2

):

frac1lambda_1 = R(1)^2left(frac11^2 - frac12^2right) = Rleft(1 - frac14right) = frac3R4

Second line of Balmer series for

He^+

(

Z=2, n_1=2, n_2=4

):

frac1lambda_2 = R(2)^2left(frac12^2 - frac14^2right) = 4Rleft(frac14 - frac116right) = 4Rleft(frac316right) = frac3R4

Since

frac1lambda_1 = frac1lambda_2

, their wavelengths (and therefore frequencies) are exactly the same. Statement II is true. ### Step 1: Final Conclusion Both statements are correct. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

Q51 jee_main_2026_21_jan_evening Hydrogen Spectrum and Energy Levels

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:

A. $(1) \ D < C < A < B$ $(1) \ D < C < A < B$
B. $(2) \ A < B < C < D$ $(2) \ A < B < C < D$
C. $(3) \ C < D < B < A$ $(3) \ C < D < B < A$
D. $(4) \ D < A < C < B$ $(4) \ D < A < C < B$

Solution

### Related Formula

Delta E = 13.6 Z^2 left(frac1n_1^2 - frac1n_2^2right) text eV

### Core Logic Let's find the values of

n_1

and

n_2

for each transition: - (A) Paschen (

1^st

line):

n_1 = 3, n_2 = 4

- (B) Balmer (

2^nd

line):

n_1 = 2, n_2 = 4

- (C) Paschen (

3^rd

line):

n_1 = 3, n_2 = 6

- (D) Bracket (

4^th

line):

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

, 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

Q54 jee_main_2026_22_january_morning Bohr Model Energy Calculations

The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ____

Jtext mol^-1

. Given:

R_H = 2.18 times 10^-11text ergs

.

A. $1.635 times 10^-18$ $1.635 \times 10^{-18}$
B. $9.835 times 10^5$ $9.835 \times 10^{5}$
C. $9.835 times 10^12$ $9.835 \times 10^{12}$
D. $1.635 times 10^-11$ $1.635 \times 10^{-11}$

Solution

### Related Formula

E_n = -R_H times fracZ^2n^2

Delta E = R_H Z^2 left( frac1n_1^2 - frac1n_2^2 right)

### Core Logic Given

R_H = 2.18 times 10^-11text ergs

. Convert this to Joules:

1text Joule = 10^7text ergs implies R_H = 2.18 times 10^-18text J

. Calculate energy difference per atom:

Delta E = 2.18 times 10^-18 times 1^2 left[ frac11^2 - frac12^2 right]

Delta E = 2.18 times 10^-18 times left( 1 - frac14 right) = 2.18 times 10^-18 times frac34

Delta E = 1.635 times 10^-18text Joule/atom

### Step 1: Conversion to per mole To find the energy per mole, multiply by Avogadro's number (

N_A = 6.02 times 10^23

):

Delta E_mole = 1.635 times 10^-18 times 6.02 times 10^23text Joule/mole

Delta E_mole = 9.84 times 10^5text Joule/mole approx 9.835 times 10^5text J mol^-1

### Pattern Recognition Energy gaps in Hydrogen:

1 rightarrow 2

transition is exactly

frac34

of the ionization energy. Watch out for per atom vs per mole unit traps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

Q64 jee_main_2026_22_january_evening Balmer Series Energy Transitions

The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is:

A. $x^2$ $x^2$
B. $fracx1.35$ $\frac{x}{1.35}$
C. $2x$ $2x$
D. $1.35x$ $1.35x$

Solution

### Related Formula

Delta E = 13.6 Z^2 left(frac1n_1^2 - frac1n_2^2right)text eV

### Core Logic Step 1: First Balmer line (

n_1 = 2, n_2 = 3

):

Delta E_1 = x = 13.6 left(frac12^2 - frac13^2right) = 13.6 left(frac14 - frac19right) = 13.6 left(frac536right)

Step 2: Second Balmer line (

n_1 = 2, n_2 = 4

):

Delta E_2 = 13.6 left(frac12^2 - frac14^2right) = 13.6 left(frac14 - frac116right) = 13.6 left(frac316right)

Step 3: Ratio of energies:

fracDelta E_2x = fracfrac316frac536 = frac3 times 3616 times 5 = frac2720 = 1.35

Delta E_2 = 1.35 x

### Pattern Recognition Sees: Ratio of hydrogen spectrum spectral line energies. Shortcut:

(3/16) / (5/36) = 27/20 = 1.35

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Physics: Atoms

Q51 jee_main_2026_23_january_morning Bohr Model for Hydrogen Like Species

Which of the following statements regarding the energy of the stationary state is true in the following one-electron system?

A. $-1.09 times 10^-18text J for second orbit of H atom.$
B. $+2.18 times 10^-18text J for second orbit of He^+text ion$
C. $+8.72 times 10^-18text J for first orbit of He^+text ion$
D. $-2.18 times 10^-18text J for third orbit of Li^2+text ion$

Solution

### Related Formula

E_n = -2.18 times 10^-18 fracZ^2n^2 text J/atom

### Core Logic Evaluate the energy of the stationary state for the given species by substituting the atomic number

Z

and the orbit number

n

into the energy formula for hydrogen-like species. ### Step 1: Calculation for Li2+ Ion For the

3^textrd

orbit of the

Li^2+

ion, we have

Z = 3

(since lithium has 3 protons) and

n = 3

.

E_3 = -2.18 times 10^-18 times frac3^23^2

E_3 = -2.18 times 10^-18 text J

### Pattern Recognition When

Z = n

, the

Z^2/n^2

ratio becomes

1

, immediately resulting in the ground state energy of a hydrogen atom (

-2.18 times 10^-18

J). This is a common shortcut for identifying correct energy states. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

Given, (A) $n = 5, m_l = -1$ $n = 5, m_{l} = -1$ (B) $n = 3, l = 2, m_l = -1, m_s = +frac12$ The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are:

Solution & Explanation

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Given, (A) n = 5, m_l = -1$n = 5, m_{l} = -1$ (B) n = 3, l = 2, m_l = -1, m_s = +frac12$n = 3, l = 2, m_{l} = -1, m_{s} = +\frac{1}{2}$ The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are:

Solution & Explanation

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Given, (A) $n = 5, m_l = -1$ $n = 5, m_{l} = -1$ (B) $n = 3, l = 2, m_l = -1, m_s = +frac12$ The maximum number of electron(s) in an atom that can have the quantum numbers as given in (A) and (B) respectively are: