Q3427. A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz. The wavelength of the electromagnetic radiation emitted by the transmitter is (c = 3.0 × 10⁸ m/s)
Year: NEET 2021
✅ Correct Answer: 219.3 m
🧠 Solution:
Frequency (ν) = 1368 kHz = 1.368 × 10⁶ Hz
Speed of light, c = 3 × 10⁸ m/s
Wavelength λ = c / ν = 3 × 10⁸ ÷ 1.368 × 10⁶ = 219.3 m
Q3475. Which one is a wrong statement?
Year: NEET 2018
✅ Correct Answer: The electronic configuration of N atom is 1s² 2s² 2pₓ² 2p_y¹ 2p_z¹
🧠 Solution:
According to Hund’s rule, pairing of electrons in orbitals of a subshell does not take place until all orbitals are singly occupied with parallel spins.
Correct configuration: 1s² 2s² 2pₓ¹ 2p_y¹ 2p_z¹.
Hence, option (a) is incorrect.
Q3448. Which one is a wrong statement?
Year: NEET 2018
✅ Correct Answer: The electronic configuration of N atom is 1s² 2s² 2pₓ² 2p_y¹ 2p_z¹
🧠 Solution:
According to Hund’s rule, pairing of electrons in orbitals of a subshell does not take place until all orbitals are singly occupied with parallel spins.
Correct configuration: 1s² 2s² 2pₓ¹ 2p_y¹ 2p_z¹.
Hence, option (a) is incorrect.
Q3429. Which one is the wrong statement?
Year: NEET 2017
✅ Correct Answer: The energy of 2s-orbital is less than 2p-orbital in hydrogen-like atoms
🧠 Solution:
In hydrogen-like atoms, energy depends only on principal quantum number n, not on l. Hence, 2s and 2p orbitals have equal energy. Therefore, statement (d) is incorrect.
Q3478. What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1, m = −1
Year: NEET 2013
✅ Correct Answer: 2
🧠 Solution:
n = 3, l = 1, m = −1 corresponds to the 3pₓ orbital.
A single orbital can have maximum 2 electrons (with opposite spins).
Q3451. What is the maximum number of electrons that can be associated with the following set of quantum numbers? n = 3, l = 1, m = −1
Year: NEET 2013
✅ Correct Answer: 2
🧠 Solution:
n = 3, l = 1, m = −1 corresponds to the 3pₓ orbital.
A single orbital can have maximum 2 electrons (with opposite spins).
Q3434. The speed of light is 3 × 10⁸ m/s. Which value is closest to the wavelength in nm of light with frequency 6 × 10¹⁵ s⁻¹?
Q3430. How many electrons can fit in the orbital for which n = 3 and l = 1?
Year: NEET (Phase II) 2016
✅ Correct Answer: 2
🧠 Solution:
An orbital can accommodate a maximum of two electrons with opposite spins. Hence, 2 electrons can fit.
Q3476. Two electrons occupying the same orbital are distinguished by
Year: NEET (Phase I) 2016
✅ Correct Answer: Spin quantum number
🧠 Solution:
Two electrons in the same orbital have opposite spins (+½ and −½). Hence, they are distinguished by their spin quantum number.
Q3449. Two electrons occupying the same orbital are distinguished by
Year: NEET (Phase I) 2016
✅ Correct Answer: Spin quantum number
🧠 Solution:
Two electrons in the same orbital have opposite spins (+½ and −½). Hence, they are distinguished by their spin quantum number.
Q3474. Orbital having 3 angular nodes and 3 total nodes is
Year: NEET (Odisha) 2019
✅ Correct Answer: 4f
🧠 Solution:
Angular node (l) = 3
Total node = radial node + angular node
3 = (n − l − 1) + l
⇒ n − 1 = 4 − 1 = 3
Hence, orbital having 3 angular nodes and 3 total nodes is 4f (n = 4, l = 3).
Q3447. Orbital having 3 angular nodes and 3 total nodes is
Year: NEET (Odisha) 2019
✅ Correct Answer: 4f
🧠 Solution:
Angular node (l) = 3
Total node = radial node + angular node
3 = (n − l − 1) + l
⇒ n − 1 = 4 − 1 = 3
Hence, orbital having 3 angular nodes and 3 total nodes is 4f (n = 4, l = 3).
Q3428. In hydrogen atom, the de-Broglie wavelength of an electron in the second Bohr orbit is (Given Bohr radius a₀ = 52.9 pm)
Year: NEET (Odisha) 2019
✅ Correct Answer: 211.6 × π pm
🧠 Solution:
For hydrogen atom, λₙ = 2πrₙ = 2πn²a₀ / Z
For n = 2, Z = 1 ⇒ λ = 2π(2²)a₀ = 8πa₀ = 211.6 × π pm
Q3450. Which is the correct order of increasing energy of the listed orbitals in the atom of titanium?
Year: CBSE AIPMT 2015
✅ Correct Answer: 3s 3p 3d 4s
🧠 Solution:
According to Aufbau principle, orbitals fill in the order of increasing (n + l) value.
Hence: 3s < 3p < 3d < 4s.
Q3432. The angular momentum of an electron in d-orbital is equal to
Year: CBSE AIPMT 2015
✅ Correct Answer: √6h/2π
🧠 Solution:
Angular momentum = √l(l+1) × h/2π
For d-orbital, l = 2 ⇒ √2(3)h/2π = √6h/2π
Q3431. The number of d-electrons in Fe²⁺ (Z = 26) is not equal to the number of electrons in which one of the following?
Year: CBSE AIPMT 2015
✅ Correct Answer: p-electrons in Cl (Z = 17)
🧠 Solution:
Fe²⁺ → [Ar]3d⁶ 4s⁰ → 6 d-electrons
Mg → 1s² 2s² 2p⁶ 3s² → 6 s-electrons
Cl → 1s² 2s² 2p⁶ 3s² 3p⁵ → 11 p-electrons
Hence, the number of d-electrons in Fe²⁺ (6) is not equal to p-electrons in Cl (11).
Q3433. Calculate the energy in joule corresponding to light of wavelength 45 nm (h = 6.63 × 10⁻³⁴ J·s, c = 3 × 10⁸ m/s).
Q3482. Which of the following is not a permissible arrangement of electrons in an atom?
Year: CBSE AIPMT 2009
✅ Correct Answer: n=3, l=2, m=−3, s=−½
🧠 Solution:
For l = 2, m can take values −2, −1, 0, +1, +2. Hence m = −3 is not permissible.
Q3456. Maximum number of electrons in a subshell of an atom is determined by the following:
Year: CBSE AIPMT 2009
✅ Correct Answer: 4l + 2
🧠 Solution:
Maximum number of electrons in subshell = 2(2l + 1) = 4l + 2.
Q3455. Which of the following is not a permissible arrangement of electrons in an atom?
Year: CBSE AIPMT 2009
✅ Correct Answer: n=3, l=2, m=−3, s=−½
🧠 Solution:
For l = 2, m can take values −2, −1, 0, +1, +2. Hence m = −3 is not permissible.
Q3414. The energy absorbed by each molecule (A₂) of a substance is 4.4 × 10⁻¹⁹ J and bond energy per molecule is 4.0 × 10⁻¹⁹ J. The kinetic energy of the molecule per atom will be
Year: CBSE AIPMT 2009
✅ Correct Answer: 2.0 × 10⁻²⁰ J
🧠 Solution:
Kinetic energy (K.E.) of molecule = energy absorbed − bond energy
= (4.4 × 10⁻¹⁹) − (4.0 × 10⁻¹⁹) = 0.4 × 10⁻¹⁹ J
K.E. per atom = (0.4 × 10⁻¹⁹) ÷ 2 = 2.0 × 10⁻²⁰ J.
Q3436. If uncertainty in position and momentum are equal, then uncertainty in velocity is
Year: CBSE AIPMT 2008
✅ Correct Answer: h / (4πm)
🧠 Solution:
From Heisenberg uncertainty principle: Δx × Δp ≥ h / 4π
Given Δx = Δp and Δp = mΔv ⇒ (Δp)² = h / 4π ⇒ Δv = h / (4πm)
Q3435. The measurement of electron position is associated with an uncertainty in momentum equal to 1 × 10⁻⁸ g·cm·s⁻¹. The uncertainty in electron velocity is (mass of electron = 9 × 10⁻²⁸ g)
Q3457. Which of the following sets of quantum numbers is not possible?
Year: CBSE AIPMT 2007
✅ Correct Answer: (1,0,−1,+½)
🧠 Solution:
For a given n, l ranges from 0 to (n−1). For n=1, l cannot be −1, so set (iv) is invalid.
Q3458. The orientation of an atomic orbital is governed by
Year: CBSE AIPMT 2006
✅ Correct Answer: Magnetic quantum number
🧠 Solution:
Magnetic quantum number (m) describes the orientation of an orbital in space.
Q3437. Given, the mass of electron is 9.11 × 10⁻³¹ kg, Planck's constant is 6.626 × 10⁻³⁴ J•s, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is
Year: CBSE AIPMT 2006
✅ Correct Answer: 5.79 × 10⁸ m/s
🧠 Solution:
By Heisenberg’s uncertainty principle:
Δx × Δp ≥ h / 4π
Δx × mΔv ≥ h / 4π
Δv ≥ h / (4πmΔx)
Given: Δx = 0.1 Å = 0.1 × 10⁻¹⁰ m, m = 9.11 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J•s
So, Δv = (6.626 × 10⁻³⁴) / (4 × 3.14 × 9.11 × 10⁻³¹ × 0.1 × 10⁻¹⁰)
= 5.79 × 10⁸ m/s
Q3415. The energy of second Bohr orbit of the hydrogen atom is −328 kJ mol⁻¹. Hence, the energy of fourth Bohr orbit would be
Year: CBSE AIPMT 2005
✅ Correct Answer: −82 kJ mol⁻¹
🧠 Solution:
Eₙ = −1312 / n² kJ mol⁻¹
Given E₂ = −328 kJ mol⁻¹, so for n = 4:
E₄ = −1312 / 4² = −82 kJ mol⁻¹.
Q3416. The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (Given ionisation energy of H = 2.18 × 10⁻¹⁸ J atom⁻¹ and h = 6.625 × 10⁻³⁴ J·s)
Q3438. The value of Planck's constant is 6.63 × 10⁻³⁴ J•s. The velocity of light is 3.0 × 10⁸ m/s. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of 8 × 10¹⁴ s⁻¹?
Year: CBSE AIPMT 2003
✅ Correct Answer: 4 × 10¹ nm
🧠 Solution:
We know, c = νλ → λ = c / ν
= (3 × 10⁸ m/s) / (8 × 10¹⁴ s⁻¹)
= 0.375 × 10⁻⁶ m = 3.75 × 10⁻⁷ m = 4 × 10¹ nm
Q3417. In hydrogen atom, energy of first excited state is −3.4 eV. Then, kinetic energy of same orbit of hydrogen atom is
Year: CBSE AIPMT 2002
✅ Correct Answer: +3.4 eV
🧠 Solution:
Total energy (E) = K.E. + P.E.
E = −K.E.
Thus, K.E. = −E = +3.4 eV.
Q3459. The following quantum numbers are possible for how many orbitals? n = 3, l = 2, m = +2
Year: CBSE AIPMT 2001
✅ Correct Answer: 1
🧠 Solution:
For l = 2, m = +2 corresponds to only one of the five 3d orbitals. Hence, only one orbital is possible.
Q3439. The energy of photon is given as E/atom = 3.03 × 10⁻¹⁹ J•atom⁻¹. Then the wavelength (λ) of the photon is (Given, h = 6.63 × 10⁻³⁴ J•s, c = 3.00 × 10⁸ m/s)
Year: CBSE AIPMT 2000
✅ Correct Answer: 656 nm
🧠 Solution:
E = hν = hc / λ
λ = hc / E = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (3.03 × 10⁻¹⁹)
= 6.56 × 10⁻⁷ m = 656 nm
Q3460. Which of the following configurations is correct for iron?
Year: CBSE AIPMT 1999
✅ Correct Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s¹
🧠 Solution:
For Fe (Z = 26): configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s² → rearranged to [Ar] 3d⁶ 4s².
Q3441. The uncertainty in momentum of an electron is 1 × 10⁻²⁵ kg•m/s. The uncertainty in its position will be (Given, h = 6.62 × 10⁻³⁴ J•s)
Year: CBSE AIPMT 1999
✅ Correct Answer: 5.27 × 10⁻⁹ m
🧠 Solution:
According to Heisenberg’s principle: Δx × Δp ≥ h / 4π
Δx = h / (4πΔp) = (6.62 × 10⁻³⁴) / (4 × 3.14 × 1 × 10⁻²⁵)
= 5.27 × 10⁻⁹ m
Q3440. The de-Broglie wavelength of a particle with mass 1 g and velocity 100 m/s is
Year: CBSE AIPMT 1999
✅ Correct Answer: 6.63 × 10⁻²⁹ m
🧠 Solution:
By de-Broglie equation: λ = h / mv
= (6.63 × 10⁻³⁴) / (10⁻³ × 100)
= 6.63 × 10⁻²⁹ m
Q3418. Who modified Bohr's theory by introducing elliptical orbits for electron path?
Year: CBSE AIPMT 1999
✅ Correct Answer: Sommerfeld
🧠 Solution:
Sommerfeld modified Bohr’s theory and introduced elliptical orbits in addition to circular orbits.
Q3442. The position of both, an electron and a helium atom is known within 1.0 mm. Further, the momentum of the electron is known within 5.0 × 10⁻²⁶ kg•m/s. The minimum uncertainty in the measurement of the momentum of the helium atom is
Year: CBSE AIPMT 1998
✅ Correct Answer: 5.0 × 10⁻²⁶ kg•m/s
🧠 Solution:
By Heisenberg’s uncertainty principle, Δx × Δp ≥ h / 4π.
When Δx is same for both particles and Δp for electron is 5.0 × 10⁻²⁶ kg•m/s, then Δp for helium atom is also 5.0 × 10⁻²⁶ kg•m/s.
Q3419. Bohr radius for the hydrogen atom (n = 1) is approximately 0.530 Å. The radius for the first excited state (n = 2) is (in Å)
Year: CBSE AIPMT 1998
✅ Correct Answer: 2.12
🧠 Solution:
rₙ = n²a₀ / Z
For H, Z = 1
r₂ = (2²)(0.530) = 2.12 Å.
Q3461. The electronic configuration of gadolinium (Z = 64) is
Year: CBSE AIPMT 1997
✅ Correct Answer: [Xe] 4f⁷ 5d¹ 6s²
🧠 Solution:
Gd (64) → [Xe] 4f⁷ 5d¹ 6s² due to half-filled f-shell stability.
Q3462. The orbitals are called degenerate when
Year: CBSE AIPMT 1996
✅ Correct Answer: They have different wave functions but same energy
🧠 Solution:
Orbitals with the same energy but different orientations (e.g., 3dₓᵧ, 3d_yz, 3d_z²) are degenerate.
Q3443. The momentum of a particle having a de-Broglie wavelength of 10⁻⁷ m is (Given, h = 6.625 × 10⁻³⁴ J•s)
Year: CBSE AIPMT 1996
✅ Correct Answer: 6.625 × 10⁻²⁷ kg•m/s
🧠 Solution:
By de-Broglie relation: λ = h / p ⇒ p = h / λ
= (6.625 × 10⁻³⁴) / (10⁻⁷) = 6.625 × 10⁻²⁷ kg•m/s
Q3444. Uncertainty in position of an electron (mass = 9.1 × 10⁻²⁸ g) moving with a velocity of 3 × 10⁴ cm/s accurate up to 0.001% will be (h = 6.626 × 10⁻²⁷ erg•s)
Year: CBSE AIPMT 1995
✅ Correct Answer: 1.93 cm
🧠 Solution:
By Heisenberg’s uncertainty principle: Δx × Δv = h / (4πm)
Δv = 0.001% of 3 × 10⁴ = 0.3 cm/s
Δx = (6.626 × 10⁻²⁷) / (4 × 3.14 × 9.1 × 10⁻²⁸ × 0.3) = 1.93 cm
Q3420. The radius of hydrogen atom in the ground state is 0.53 Å. The radius of Li²⁺ ion (Z = 3) in a similar state is
Year: CBSE AIPMT 1995
✅ Correct Answer: 0.17 Å
🧠 Solution:
rₙ = n²a₀ / Z
For H: Z = 1 → r₁ = 0.53 Å
For Li²⁺: Z = 3 → r₁ = 0.53 / 3 = 0.17 Å.
Q3464. For which one of the following sets of four quantum numbers, an electron will have the highest energy?
Year: CBSE AIPMT 1994
✅ Correct Answer: 4, 2, −1, +½
🧠 Solution:
According to (n + l) rule: higher (n + l) → higher energy.
4d (n + l = 6) > 3d, 4p, 5s.
Q3463. If an electron has spin quantum number +½ and magnetic quantum number −1, it cannot be present in
Year: CBSE AIPMT 1994
✅ Correct Answer: s orbital
🧠 Solution:
For s-orbital, l = 0 and m = 0 only. Hence m = −1 is not possible.
Q3446. The electron was shown experimentally to have wave properties by
Year: CBSE AIPMT 1994
✅ Correct Answer: Davisson and Germer
🧠 Solution:
The wave nature of the electron was proved by Davisson and Germer experiment. The diffraction pattern of electrons was found similar to X-rays, confirming their wave character.
Q3445. In the photoelectron emission, the energy of the emitted electron is
Year: CBSE AIPMT 1994
✅ Correct Answer: smaller than the incident photon
🧠 Solution:
In photoelectric effect, the energy of the emitted electron is smaller than that of the incident photon because some energy of the photon is used to eject the electron, and the remaining appears as kinetic energy.
Q3421. When an electron of charge e and mass m moves with a velocity v about the nuclear charge Ze in circular orbit of radius r, the potential energy of the electron is given by
Year: CBSE AIPMT 1994
✅ Correct Answer: −2Ze² / 4πε₀r
🧠 Solution:
Potential energy = Work done in bringing charge = −2Ze² / 4πε₀r.
Q3412. Which of the following is never true for cathode rays?
Year: CBSE AIPMT 1994
✅ Correct Answer: They are electromagnetic waves
🧠 Solution:
Cathode rays are not electromagnetic waves because they do not have electric and magnetic components perpendicular to each other.
Q3422. If ionisation potential for hydrogen atom is 13.6 eV, then ionisation potential for He⁺ will be
Year: CBSE AIPMT 1993
✅ Correct Answer: 54.4 eV
🧠 Solution:
Ionisation energy ∝ Z²
For H: Z = 1 → 13.6 eV
For He⁺: Z = 2 → 13.6 × 4 = 54.4 eV.
Q3465. Electronic configuration of calcium atom can be written as
Year: CBSE AIPMT 1992
✅ Correct Answer: [Ar] 4s²
🧠 Solution:
Ca (Z = 20): [Ar] 4s².
Q3423. The energy of an electron in the nth Bohr orbit of hydrogen atom is
Year: CBSE AIPMT 1992
✅ Correct Answer: −13.6 / n² eV
🧠 Solution:
Eₙ = −13.6 / n² eV for hydrogen atom (Z = 1).
Q3469. In a given atom no two electrons can have the same values of all the four quantum numbers. This is called
Year: CBSE AIPMT 1991
✅ Correct Answer: Pauli’s exclusion principle
🧠 Solution:
According to Pauli’s exclusion principle, no two electrons in an atom can have the same values of all four quantum numbers.
Example:
For 1s²:
First electron: n = 1, l = 0, m = 0, s = +½
Second electron: n = 1, l = 0, m = 0, s = –½
Hence, spin quantum number must differ.
Q3468. For azimuthal quantum number l = 3, the maximum number of electrons will be
Year: CBSE AIPMT 1991
✅ Correct Answer: 14
🧠 Solution:
When azimuthal quantum number (l) = 3,
Number of orbitals = 2l + 1 = 2×3 + 1 = 7
Each orbital can contain 2 electrons.
Total electrons = 7 × 2 = 14
Alternatively, total electrons = 4l + 2 = 4×3 + 2 = 14
Q3467. The order of filling of electrons in the orbitals of an atom will be
Year: CBSE AIPMT 1991
✅ Correct Answer: 4s, 3d, 4p, 5s, 4d
🧠 Solution:
The sequence of energy level can be remembered by the diagram showing the order of increasing energy levels.
Hence, the correct order is:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f, 5d, 6p, 7s...
Q3466. The electronic configuration of Cu (Z = 29) is
Year: CBSE AIPMT 1991
✅ Correct Answer: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰
🧠 Solution:
Due to exchange energy and symmetry, Cu has configuration [Ar] 3d¹⁰ 4s¹.
Q3470. The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
Year: CBSE AIPMT 1990
✅ Correct Answer: 6
🧠 Solution:
When n = 2 and l = 1, the subshell is 2p.
Number of orbitals in p-subshell = 2l + 1 = 3
Each orbital can hold 2 electrons.
Total electrons = 3 × 2 = 6
Q3472. Number of unpaired electrons in N⁺² is/are
Year: CBSE AIPMT 1989
✅ Correct Answer: 1
🧠 Solution:
Electronic configuration of N = 1s² 2s² 2p³
N²⁺ means removal of two electrons → 1s² 2s² 2p¹
Hence, there is only one unpaired electron.
Q3471. The maximum number of electrons in a subshell is given by the expression
Year: CBSE AIPMT 1989
✅ Correct Answer: 4l + 2
🧠 Solution:
Number of orbitals in a subshell = 2l + 1
Each orbital can hold 2 electrons.
Therefore, total electrons = 2(2l + 1) = 4l + 2
Q3424. Which of the following statements do not form a part of Bohr’s model of hydrogen atom?
Year: CBSE AIPMT 1989
✅ Correct Answer: The position and velocity of electrons in the orbit cannot be determined simultaneously
🧠 Solution:
Bohr's model allows fixed orbits and quantised energy levels. Statement (d) is from Heisenberg's uncertainty principle, not Bohr's theory.
Q3473. The number of spherical nodes in 3p-orbital is/are
Year: CBSE AIPMT 1988
✅ Correct Answer: one
🧠 Solution:
Number of spherical (radial) nodes = n – l – 1
For 3p orbital, n = 3 and l = 1
Nodes = 3 – 1 – 1 = 1
Q3426. The spectrum of helium is expected to be similar to that of
Year: CBSE AIPMT 1988
✅ Correct Answer: Li⁺
🧠 Solution:
The spectrum of an atom depends on the number of electrons present in it. Helium has two electrons, so the spectrum of Li⁺ (Z = 3) is similar to that of helium because both He and Li⁺ have two electrons.
Q3425. If r₁ is the radius of the first orbit, the radius of nth orbit of H-atom is given by